PHYSICS TEST 2 2025 FULL SOLUTIONS

Ladies and gentlemen, welcome to today's session. in this video we are going to talk about physics questions. Now this is test two stream two and this paper has got interesting questions. So let's pay attention and let's see how best we can we can answer these questions. the most important thing that want you guys to understand is when you want to pass physics here are the techniques that you're supposed to use. One understand the question. Two come up with data. Three know the formula that you supposed to use. So in physics as long as you don't understand question especially people who stays across you know people who stays across it's difficult for them to get this concept concept because as they are moving from from campus going across then they'll be like no let's check something here. So people from across listen and listen very carefully understand the question read the question understand it come up with data know the formula that we're going to use then as you are preparing for your exams want you guys to focus on this focus on the test you wrote both test one test two both stream one stream two and also focus on the quizzes you have been writing and then you need to solve all the questions from tuto sheet once you solve all the questions from tuto throw sheet the test you have been writing then at least you are good to go then you can also go through the pass paper for 2023 and 2024 the exam pass paper okay again another thing is there are some questions that sometime they get direct from the lecture notes especially from those lecture notes from the course coordinator so try to check those those two and example that they give you in the lecture notes. So if you deal with these these things that I've mentioned here believe me it's going to be easier for you. So without wasting much of your time let's begin our discussion with the first question. So the first question is under relative velocity and this is one of the simplest topic in physics. Okay. You drive north on straight two lane road at constant speed of 888 km/ hour. truck in the other end approaches you. Meaning that you are moving in the other direction. The truck is again moving in the other direction at constant speed of 104 km/h. Find the relative find the truck's relative velocity or velocity relative to you. Now let me let me just explain something here in case you don't understand. So we here is car this car is moving due north. So this car is the velocity of you which is 88 km/ hour. Then now since you are moving toward positive x-axis this is going to be positive. Then another car here here is the road. This is the road. Another car is coming is approaching you meaning it's moving in opposite direction. You understand my point? So this is the velocity of the truck. So the truck now has got the velocity of now since the truck is moving in the opposite direction that velocity will be negative. What velocity have we been given? We've been given 104. So here is 104ative 104 kilometers hour. Now the question is they want us to find the velocity relative to you. Okay. So the velocity of the track okay relative to you or you can say relative to you. You put it like this is given by the velocity of the track minus the velocity of view. That's the basic idea. So if you have got two object let's say you have got velocity of and velocity of then you want to find the velocity of relative to So the velocity of relative to is given by VA minus VB. Then the velocity of relative to which is VBA is going to be given by BB minus VA. That's the basic idea. So here we want to find the velocity of the track relative to you. Meaning that is the velocity of the track relative to you. That's what I've done. In other books, they are not going to put these to my things. They just say vtu. Now is the track. This is the velocity of the truck relative to you. So it's the velocity of the truck minus the velocity of you. So let's now see what is the velocity of the truck. isative104 minus what is the velocity of the or the velocity of is 88. Now the answer that we're going to get is going to be negative. But what does this negative tells us? It's velocity because velocity is vector quantity remember. So -192. Now we get the magnitude. So the magnitude this negative is just telling us where it is going now. meaning that it was going toward the south. Okay, because it's negative meaning it was going down. So the direction is down and the answer is 192 km/ hour south you have seen right? So that's the basic idea behind these things. It's not it's not something that is complicated if you look at it. It's not okay. So these guys as long as they approaching each other the other one will be positive the other one will be negative in the same line. If they are going in the same direction it will have the same sign. Okay that's the basic idea. So now here the best option is 192 km/ hour due south. So that's the answer. Okay let's now go to question two. So question two is saying again it's the same thing we're talking about relative velocity still car travels east with speed of 50 km hour rain drops after falling now let me just before answer this sometimes you might be wondering guys to say but what's happening it's like this have you come across situation where you are with maybe bike or anything just and then or Maybe you are running, the other person is running, then that person, you pass that person, that person also passes you. Or when you're in car and then sometimes you're going to discover to say even the the way the cars are moving like you like how how is it moving just from here. So this this means that in just per hour this guy is going to move at 192 km with distance of 192 km. That's what wanted to clarify. So car traveling east due east with speed of 50 km/h rain drops are falling at constant speed vertically with respect to positive or respect to the earth. The traces of the rain on the on the side window of the car makes an angle of 60° with the vertical with the vertical. with the vertical find the velocity of the rain with respect to the earth is basically the same question is similar to the question we from solving so you realize that nothing is going to be difficult here again the time we're talking about stream one it's the same thing where we're talking about tan theta think you can remember very well so you have got triangle and you have been given you want to in this case we want to find it's like we want this remember what we did in in stream one don't Don't waste much time where we say so in that case it's going to be 50 that we have been given and we want to find the other component which is the which we don't know that's the velocity we want to find okay so in that case that's what we need to know so we want that component for us to find this component now what we're going to do is it's the same as to find the to be 58 * 10 and the angle we have been given 60. So if you punch on your calculator and you remember since the rain is dropping going this side from the question here which is saying car travels due east with speed of 50 km hour. Rain drops are falling at constant speed vertically with respect to the earth. The traces of the rain on the side window of the car makes an angle of 60° with the vertical. Find the vertical find the velocity of the rain with respect to the earth. So definitely the answer should be supposed to have downward because the rain will be dropping down just like that. But we want this vertical component. So it will be downward. But the answer now using that concept that we did if you can remember very well we talked about that triangle we said when another one is going in this direction another one in this direction then we had this starting from there there okay so remember what we talked about the angles done is opposite over hypotenus so that's what we're going to use use the same principle so then it's going to be tan theta is equal 50 / then is 50 / tan theta. So what happens is 50 tan sist. So if you punch on your calculator again explore more on relative velocity it's also going to help you lot. It's 28 20 28.86. So it has to be due north due south which is downward. So the reason why I'm not going to waste much time to explain about this question is basically the same as the question is similar to the question we were talking about in stream one. So try to watch that video for stream one so that you understand better as well. Let's now go to question three because of time. If if man weighs if man weighs 9 900 Newtons on earth, what would be the weight? What would be the way? What would he where on Jupiter where the acceleration due to gravity is that? Now look at this. The acceleration due to gravity. The acceleration due to gravity it can change depending on the planet you are in and then the mass is not going to change. So now we know that weight is given by the mass time on is 9.8 so weight is 9 900 the mass we don't know then this is 9.8 so we divide both sides by 9.8 8 here 9.8. So the mass will not change and we end up having 900 divided by 9.8 that is 91.84 kgs. So what happens guys is this now we want to find the weight Jupiter. So it's the same mg but is the same. has now changed 25.9. So 25.9 * that I'm getting 2 newtons not newtons. So this is the answer but all answers are in kilon. So divide it by,000 that gives me 2 33 which is 38 kon newtons. So is the correct answer. Now let's go to part question four sorry. Three identical balls are thrown from the top of the building or with the same initial speed. The first ball is thrown horizontally. The second at an angle above the horizontal and the third at some angle below the horizontal negating air resistance. Rank the speed of the ball. So as long as you are throwing anything down here is building. can throw something going here. throw something going there. Another one with just an angle just here. One thing want you to understand is that these guys are going to be moving with the same velocity. One, they'll moving with the same velocity. They reach on the ground at the same time because there's no air resistance. They're going to hit the ground at the same time. Then at the same time, yeah. So all these guys are moving at the same they're moving at the same with the same velocity again they're going to hit on the ground at exactly same time. So there's no need of panicking on this question. So the best answer is and it's going to be So all the three strike the ground at the same speed and at the same time as well. So question five, bus again this question. Okay, this question is the same as the one we solved in stream one. So think it's going to be easier. It's the same formula. So you are finding the net force bus of 100 1,000 kg increases speed from from 100 that is initial to 200 in 60 seconds. They've just changed figures. So they want you to find the Fnet. told you that the FN is given by mass time acceleration. We don't have acceleration. So the acceleration is final minus initial divided by time. So the mass is a,000. They have not changed. Even in stream one, it was the same. The final is 200 minus 100. Then divide it by. So then it's going to be a,000. 200 - 100 is 100. 100 divided by so in short 0 and 0 will go so say 10 10 / 6 10 / 6 that is 1.66 so I'll say 1.67 so you are multiplying it by th00and that gives me 1,000 so I'll put it in scientific notation because all my answers there are in scientific notation so it is actually 1.66 66 or 1.7 67 * 10 ^ 3 Newton brings me that's the correct answer again question six we talked about it but in this case this is different scenario here there's friction so there's friction now I'm going to explain here I'll take time to explain because there's friction and we didn't talk about friction in semester one In in stream one, box of mass 8 kgs rest on horizontal rough surface. string attached to the box passes over the smooth puri and support 2an 2.0 kg mass. So here is the thing when the box is released and the friction force is 6 newtons act on it. What is the acceleration of the box? One thing want you to understand is this when we're talking about friction force, py system where there is friction force like said in stream one, let's call this M2. Let's call this M1. According to the question we have been given 2 kg and 8 kgs. So here there is 8 kgs and here there is 2 kgs. So there's friction force that is acting since it is moving. So we will say it's the cam. Then since it is moving in this direction we are aware of that the weight force which is weight one is M1G. the summation of the forces the force that makes the whole system to move in this direction is the weight one but this wet one has to overcome friction so we say minus friction so want you to to to check the difference with the one we are talking about in stim where we don't have friction when we don't have friction it just ends here we just say the summation is given by one because there's no friction but now this one has to overcome friction Now because we have friction so this you still add the mass because we have got two masses. So this is m1g minus the friction force because we have the value. Divide both side by m1 m2 divide both side by m1 m_sub_2. These will cancel. is equal to M1 minus then friction force then M_sub_1 M2 let's plug in the values m_1 is 2 * 9.8 minus the friction force is 6 then this is 2 then this is 8 acceleration will be equal to what is 9 2 * 9.8 is 19.6 19.6 6 - 6 divide that by 2 + which is 10. is maybe I've made mistake. 2 * 19.6 19 2 * 9.8. Yeah, that's 19.6. That's 19.6. So 19.6 6 / 10. minus sorry minus 13.6. So 13.6 / 10 and 1.36 m/s squared. That is my acceleration. So if go back, look at the answers, my answer is 1.3. 1.36. There's no 1.36, but can round it off. We have found 1.36, which is the same as 1.4 m/s squared. So, therefore, is correct. Let us look at question seven. Two blocks of mass. Again, this one we talked about it, right? no this one is different. Okay this one is different. Two blocks of mass 2 kg and 3 kg are connected by right string and pulled by force of 15 newtons. Find the acceleration of the system. This question comes from body system tension force. So here's free body diagram. This is three kgs and there's rope in between and this is two kgs. Allow me to call this m1 and this one m2. So the force that is pushing this guy is 15 newtons. want you to understand certain things here when you've got such type of questions. The summation of the forces, the force that makes the whole system, there's no friction at all to move in this direction. It's just the According to second law, replace this with mass time acceleration. But this we have got two masses. So we need to add the mass m1 m2 * Then you have this divide it by m_sub_1 m_sub_2 m1 m_sub_2 people from across hope you are able to follow through. So the force divided by m_sub_1 m_sub_2 what is the force 15 mass 2 okay here I'll say is 3 + 2 will be 15 / 5 that is 3 m/s squar hope this is simple explained on how to deal with this if there's friction want you to check out the videos that are there on the website if you have not Get registered with us. Take time register with us and then when you register with us now you still be able to access semester 2 work. That would mean to say it's going to help you to even stud failure before you come for semester 2. So go on website if you don't know the website go on Google and type transcendent. The website is going to pop up. Register with us. If you're having any challenge to register, check the link in the description below. Okay, you'll be able to be you'll be able to to join the group, the WhatsApp group. Then believe it's going to help you. So the answer is 3 m/s. Let's jump to question eight. car of mass. This question again we solved it. It was there in 2024 final exam. solved it perfectly. hope you guys managed to follow through. So here I'm going to explain again. But if you want to understand fa 2024 paper is also there. solved it. In fact it was class. So we solved it and explained what what you are supposed to know. car of mass400 kg is traveling on straight line or horizontal road at constant speed of 25 m/s. The output power from the car's engine is 30 kilow. The car then travels up slope at 2° to the horizontal maintaining the same speed. What is the output power of the car's engine when traveling up the slope? The most important thing here to understand is the constant. This guy called constant speed. When we have got constant speed, if you have got any force, any object that is moving with constant speed, the force that is pushing that object and the friction force, they are going to be the same. Why am saying this? Because the net force is zero. Okay? So now here even if don't know the force what want you to understand is that as it was moving here this one was was moving with certain force know that power is work time this but there's also another formula I'm not going to drive it it's also force time velocity so if you've got they're talking about the force and the velocity in this case we've got the velocity I'm going to use this formula so power is 30 kilo what time a,000 that is 30,000 then the velocity is 25 so we divide it by 25 here 25 so is 1 1200 ntons this 1200 ntons is the force that was being used pushing this with constant speed at the same time when this one now started now reaching at this point going toward this what want you to understand is that again here it was still at at constant. But when we're talking about incline, there is force of gravity that naturally exist. But again, another force that is existing is since we know that this object is moving in this direction. Okay, this gravity, the force of gravity plus the force that we have found here plus 1,200 ntons is going to be the force that is going to be pushing this guy going up the incline because it is moving at constant speed. So the net force is going to be zero. What mean that the net force in this case will say the forces are the applied force that don't know this force then it's going to be minus because these ones this one is now opposing and also this one is also opposing so minus 1,2 and also fg when we're talking about incline fg is always mg sin theta so we can find fg fg the mass is400 * 9.8 then sin what? Sin 2°. So use your calculator to punch 1,500 time 9.8 * sin 2. So the answer that I'm getting is 478.82 Ntons. Now when add this, I'm going to shift this to go to the other side because the net force is zero when it is moving at constant speed. We have been told constant speed. So zero then we have got that want. This is now 1200 minus 478.82. So these ones are adding. So if these ones are adding it will be minus. If add those I'll be able to get 16 -6 78.82. Shift this to go to the other side. So that is 1678.82 ntons. That is now the force that was used to be pushing this guy now going up the incline. Now want to find the power. Okay. So power is equal to the force time in this case because have got the force and the velocity. So power the force is 1678.8 2. This is two. Then time the velocity 25. So power will be this answer time 25 that is 41 0.5 3 all the answers are in kilow divided by th00and so power is giving me 41.97 kilow. Let's round it off. That becomes kilow. So therefore is the correct answer. It's making sense, right? Yes, it does. Unless you're from across. Now let's go to question again. Question nine, we talked about it in stream one. It's just basically the same concept. The question is 5 kg mass is raised 10 So what is the weight done against the gravity? So against the gravity we're talking about gravity. This is the gravity now on earth. So against you are going toward this. So that's the FG we're talking about which is the the force of gravity. Then the distance is raised meaning that the distance is again moving at the same pace 10 So the work done will be positive. So work is equal to the force time distance. So the force of gravity in that case. So the force of gravity is given by mg * is 5. is 9.8. is 10. This is 50 or let me just say this is 40. 9.8 * 10 98. 98 * 5 490 jaws. So they had to put negative and positive but the one for negative is not correct because against you going away from gravity so it's positive unless if they ask you to say toward the gravity it was raised. So the question maybe they ask you to say it was raised it was raised 10 then what is the work done toward the gravity? Toward the gravity is going to be negative. Question 10. Question 10 is saying 3 kg broke starts from rest at the top of the 30 kilometers 30 30° incline. and slide slides distance of 2 down the incline in 1.5 seconds. Find the magnitude. Magnitude of what? Find the magnitude. Find the magnitude and acceleration of the block and the coefficient of friction for this. Okay, let's come up with free body diagram. If I'll be able to remember an angle of what? Forgotten. An angle of 30°. So it has started from rest going in this direction. But there's friction which is opposing. Remember there's no any other force that is making it to slide. Meaning that there's natural force that is FG. When we're talking about incline plane, FG is always mg sin theta and normal force is always mg cos theta. And explained this the time was explaining about Newton's laws of motion where they coming from. If you're interested, but no one is going to ask you where they coming from. Now look at this from the data that we have been given. They have given us the time they have given us the time 1.5 the distance two 1.5 seconds the distance 2 what else have we been given the mass to be 3 kg. Now our goal is to find the acceleration and the mu value which is the mu in this case. So what want you to understand is that since we don't have we started from rest meaning that the initial speed is zero. The formula that can use to find the acceleration is is equal to the initial * + half Now that the initial is zero, this is going to be cancelled. So is = half t². So cross multiply is 2d = ^ 2 / ^ 2. Even here ^ 2. So is 2d / ^ 2. So let's now find the So will be equal to 2. The is 2. So then is 1.5. square 8. So will be equal to 2 * 2 is 4 / 1.5. So I'm getting 1.7777 which is the same as 1.78 m/ second squared. Now that we have found our acceleration to be 1.78 m/s squared, let's now look at let's now talk about how we can find the the coefficient. So you use now the summation of the forces in this direction. Let's assume that this is the direction. So the summation of the forces in direction, the force that we have that is pushing the object to go in this direction is the fg. But friction is opposing. Okay. And want you to want you to remember that friction is given by the muk that is the FK the kinetic friction is FK time the normal force. So now the summation of the forces according to second row you replace it with mass time acceleration. FG is mg sin theta. That's what we have said from here. minus FK is mu the normal force the normal force is mg cos theta our goal is to find the mu value can see that there's mass everywhere can cancel mass have remained with being = sin theta minus mu cos theta shift this to go to the other side shift to go to the other side so then mu * cos theta is equal to sin theta minus divide it by cos theta even here cos theta these will cancel. Therefore the mu cm in case you want to copy that. Therefore the muk is sin theta minus / cos theta. Let's plug in the values. The mu will be is 9.8 sin 30 - 1.78. Divide this by 9.8 cos 30°. I'll do it direct. So 9.8 sin 30 the answer get is 4.9 then - 1 so on top I'm getting 3.12 down part is 9.8 8 cos 30. So what I'll do is it's it's giving me I'm decimals number 9.8 cos 30. So the answer that is giving me what I'm going to do is it's better off for me just to divide direct. Why can't just say don't want to make any mistake. So the answer that found is 3.12. divide that by 9 9.8 cos cos 30. So I'm getting my answer my mu value as 3 0 6. can round it off to three decimal places which is 0.368. So this is the mu value. So the mu value the acceleration is 1.78. The mu value 0.368. As you are solving view mu mu is always less than one. So when you're solving as long as it's greater than one failed that question. So the best answer was 1.78 is okay that marks the end of this paper. hope you guys have gotten something. hope you have enjoyed. Okay. So in case you want my number just minute in case you want my number and you you want to join in the group for the preparation of semester 2 that we're going to start you can get in touch with me using this number plus 2607 Sir Isaac Newton So thank you guys for watching my video. believe as you are watching this video you have already subscribed and you have already like the video. Okay. Cheers.