Test Papers – ICSE – Class – X
Std-11, Physics, Kinematics, Ch-05 Projectile Motion
Projectile Motion:
What is projectile motion? When a particle is projected at certain angle. Suppose this is x and this is y, and we have thrown a ball like this from here, we have thrown it u initial speed at theta angle, so this object will perform this motion. Yes or no? This is called projectile motion. And we are dealing with case 1.
It is called ground to ground projectile. It is called ground to ground projectile because the point of projection and point of landing, both are on the same horizontal level. So that is called ground to ground projection. Can I say that, see this ball has gone up and it has come down. Now add this to this. This vertical motion which we called as free fall in the previous topic, in that if we add this linear motion then what will we see. The combined effect of both and that will be known as the projectile motion. This two dimensional motion is basically the combination of this linear motion and this linear motion. And what is fascinating is that this is the motion which we already know a lot about. Yes or no?
So let’s draw the basic, we will define motion for x and y, okay. Initial velocity x, acceleration x, displacement x. Initial velocity y, acceleration y, displacement y. This is the technique of learning projectile motion by making this table first of all. We will fill in the values in this table, okay.
Now we had learnt about a topic on vectors. And on that topic what had I told you, I had told you that the most useful thing for us will be. Component. u velocity at theta angle. Can you make two components of this velocity? What will be this component, class?
u cos theta. And what will be that component?
[students answering]U sine theta.
Now let’s fill in the table. Whatever you can understand by looking at this information we will fill this table, okay. Tell me ux initial velocity along x axis what will that be? u cos theta. Initial velocity along y axis, what will that be? u sine theta. Acceleration along x, acceleration along x. There is only one acceleration working on this, which one is that g. And where is it working downward. So what is horizontal acceleration, yes or no? So ax zero. That’s critical. What is ay? g or minus g.
[minus g]In 1D what was the technique we had made to decide the sign for displacement and acceleration. According to that rule it will be minus g, yes or no? What is horizontal displacement? You don’t know. This will be the horizontal displacement. Right. So let’s call it R, which we don’t know. What is vertical displacement? What is total vertical displacement?
Std-11, Physics, Kinematics, Ch-04 More Problems on 1 D Motion
So we were dealing with this particular question that the ball is projected vertically upward with some velocity and we are looking forward to find or to calculate distance or displacement covered in last t seconds. What is the meaning of last t seconds? Last t seconds of upward journey, this is what we have written, right. We may have projected this ball at any velocity and there will be a final moment when the velocity will be zero, so final velocity zero. So total displacement during the motion. First let us understand the meaning of last t seconds. What I mean to say is that particle is in upward motion for 10 seconds, that means to reach from here to there it takes 10 seconds. So I am asking you motion of last 3 seconds. I want to know the motion of the last 3 seconds, so last t seconds, some t seconds. Let’s say to reach from here to here it takes capital T time. So from that capital T, how much displacement has it covered in the last t seconds. Done?
Suppose this is the x displacement in the last t seconds, okay. Now how to calculate that?
[Students answering]
Total time, one student is saying that there will be total time capital T and from here to here it will be small t. So what is the time of this journey? Capital T minus small t. If we minus this journey time or displacement from total time so we will get the displacement of last t seconds.
Let’s say, now we don’t know the initial speed. Let’s say initial speed is u, but we don’t know the initial speed. And total time is capital T, if total time is capital T that means after capital T time its velocity will become zero. Yes or on? So will capital T carry this relation or no? There will be this relation between capital T and initial speed. And here we can we can write u as g capital T. Class, is that clear? During the upward journey capital T. Now see we need total, and total we know that if initial speed is u what is the total displacement. Total displacement is equal to maximum height so u square by 2g. And this u square by 2g we can write it with the help of this formula as g square T square divide by 2g, and when we solve that what will we get, gT square by 2. This is total displacement. From total displacement capital T minus small t is the displacement of time. Should we minus that? So this is displacement of Capital T minus small t time. How much is that? Second equation ut plus alpha t square. So it is u, what is the time, capital T minus small t minus g by 2 capital T minus small t square. And again u can be replaced again. It will be gT. Agreed? We can simplify this. This will be gT square minus g capital T small t. We can open this, T square plus small t square minus 2 capital T small t, right. We can simplify it some more. gT square minus g capital T small t s gT square by 2 minus gtsquare by 2 small t. Minus minus becomes plus 2 will be cancelled with 2 g capital T small t.
Here see very carefully, this gets cancelled with one another and this and this can be minused. What is the final value.
Std-11, Physics, Kinematics,Ch-03 Problems on 1 D Motion
So, students, in last lecture we were dealing about linear motion with constant accelerations and we were having some problems or some practice questions on that concept. Let’s take few more examples on the same idea of linear motion with constant acceleration.
Let’s consider a problem, an air balloon is moving vertically upward with constant speed of 40 meters per second. When the height of balloon from the ground is 100 meters, the student in the balloon releases a stone. Then we need to find the trajectory of the stone and the time taken by the stone to reach the ground, okay.
See there is one idea, if an object is released from a moving system, we are not throwing, we are releasing, if an object is released from a moving system, then this object will carry the velocity of its parental system. The system from which you have released, it will carry the parental velocity but you have to notice an important thing that it will not carry acceleration. Because acceleration is dependent on contact force. The moment the student will leave the stone, no longer the force of student is acting on the block or the stone. That’s why now they are independent of each other. So there will be no acceleration sharing between them. But the stone will carry the velocity. That means just after the release the stone will have a vertical velocity of 40 meters per second, 100 meters from the ground. So under the effect of this velocity the stone will move first vertically upward, somewhere it will attain the zero velocity and then it will turn and fall back to the ground. And during this entire motion it will experience a constant acceleration, g. See, after immediate release the velocity remains the same but acceleration changes immediately. The acceleration of the stone at the start was equal to g and now it is in free fall and it is under gravity. The balloon is still moving with constant velocity. So you will get the velocity of the parental system but you will not get the acceleration. Now we have to calculate the time. We can use the second equation. Total displacement is minus 100, initial velocity is 40 in upward direction t minus half gt squared. And once we will solve this (3:26) we will get the answer t is equal to 10 seconds. So after 10 seconds it will reach the ground. The conceptual idea that is involved in this problem is that whenever an object is released from a moving system it will carry the velocity of parental system but it will not carry the acceleration. Done, okay.
Let’s move to the next problem.
Okay, let’s look at one more good problem. A particle is projected upward with initial velocity u. We will observe one certain height, like we are sitting here and watching. And this h is less than h maximum which this object will attain. Now see what the student who is observing the motion, see what he observes. It will move upward and then it will move downwards. That means at the same height this object will appear twice to us. First in the upward journey and second in the downward journey. Let’s say t1 is the initial time and t2 is the, here first or second time. We have to relate t1 and t2 and let’s say we want to find the values of t1 and t2, then what we can do. Okay, so let’s discuss about t1 and t2.
Std-11, Physics, Kinematics, Ch-02 Equation Based Problems
A particle moving with constant acceleration, moving with constant acceleration of 4 meters per second square covers 60 meters distance in 8th second of its motion. Find the initial velocity of the particle?
This is totally fine.
Displacement in 8th second will be initial speed plus acceleration by 2, 2 into 8 minus 1. Yes or no? So it is 60 = u 4 by 2, 2 into 8 minus 1. This will come out to be u plus 30, so u is equal to 30 meters per second.
Next, a particle start from rest moves with constant acceleration along a straight line if displacement of particle in t seconds is s1, in next t seconds it is s2, in next t seconds it is s3, and so on. Then find the value of s1 raised to s2 raised to s3, like that. And then find the value of this. Particle starts from 0 initial velocity moves with constant acceleration. The displacement of first t second is s1, and the displacement of the next t seconds is again s1, sorry it is s2, then s3 and so on so we need to find a relation between s1, s2, s3, s4 and so on, done.
