Solveeit Logo
Login

Question

Question: A ball is bouncing elastically with a speed of \(1\,m{s^{ - 1}}\) between walls of a railway compart...

A ball is bouncing elastically with a speed of 1 ms−11\,m{s^{ - 1}} between walls of a railway compartment of size 10 m10\,m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 ms−110\,m{s^{ - 1}} parallel to the direction of motion of the ball. As seen from the ground.

A. The direction of motion of the ball changes every 10s10s
B. Speed of ball changes every 10s10s
C. Average speed of ball over any 20s20s interval is fixed
D. The acceleration of ball is the same as from the train

Explanation

Solution

Here first we have to find the time taken by the ball to collide.
Then we have to find the average speed of the ball for the total time taken and total distance travelled.
At last we have to see whether the motion of the train is uniform or not.
Here we have to apply the concepts of motion.

Complete step by step answer: Given,
Speed of the ball =1 ms−1 = 1\,m{s^{ - 1}}
Distance of the ball from the walls =10 m = 10\,m
Velocity of the ball =10 ms−1 = 10\,m{s^{ - 1}}
Since, the motion of the train is parallel to that of the ball. So, both the direction of the ball and the train will be the same. Thus, the speed of the ball is =10+1=11 ms−1 = 10 + 1 = 11\,m{s^{ - 1}}
.After a collision the direction of the ball changes. So, the speed of the ball is =10−1=9 ms−1 = 10 - 1 = 9\,m{s^{ - 1}}
.We know that time =distancespeed = \dfrac{{{\text{distance}}}} {{{\text{speed}}}}
So, time taken by the ball for collision =101=10 s = \dfrac{{10}} {1} = 10\,s
Hence, we can say that option B is correct.
Now, from the above discussion we can see that the speed of the ball changes before and after collision. So, we have to consider two distances. Hence, the total distance will be =10+10=20 m = 10 + 10 = 20\,m
. So, the total time taken will be =20 s = 20\,s
. Hence, the average speed will be =2020=1 ms−1 = \dfrac{{20}} {{20}} = 1\,m{s^{ - 1}}
, which is the uniform speed.
Thus, option C is correct.
We find that the train is moving at a constant velocity of 10 ms−110\,m{s^{ - 1}}
. The train thus acts as a frame of reference which is the same for the ball as well.
Hence, option D is also correct.

Note: In this question, we can easily get confused between all the options. But we have to examine the entire options one by one to see if they are correct or not by applying the concepts of motion.