
A ball is bouncing elastically with a speed of $1\,m{s^{ - 1}}$ between walls of a railway compartment of size $10\,m$ in a direction perpendicular to walls. The train is moving at a constant velocity of $10\,m{s^{ - 1}}$ parallel to the direction of motion of the ball. As seen from the ground.
A. The direction of motion of the ball changes every $10s$
B. Speed of ball changes every $10s$
C. Average speed of ball over any $20s$ interval is fixed
D. The acceleration of ball is the same as from the train
Answer
568.5k+ views
Hint: 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\,m{s^{ - 1}}$
Distance of the ball from the walls $ = 10\,m$
Velocity of the ball $ = 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\,m{s^{ - 1}}$
.After a collision the direction of the ball changes. So, the speed of the ball is $ = 10 - 1 = 9\,m{s^{ - 1}}$
.We know that time $ = \dfrac{{{\text{distance}}}}
{{{\text{speed}}}}$
So, time taken by the ball for collision $ = \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$
. So, the total time taken will be $ = 20\,s$
. Hence, the average speed will be $ = \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 $10\,m{s^{ - 1}}$
. 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.
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\,m{s^{ - 1}}$
Distance of the ball from the walls $ = 10\,m$
Velocity of the ball $ = 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\,m{s^{ - 1}}$
.After a collision the direction of the ball changes. So, the speed of the ball is $ = 10 - 1 = 9\,m{s^{ - 1}}$
.We know that time $ = \dfrac{{{\text{distance}}}}
{{{\text{speed}}}}$
So, time taken by the ball for collision $ = \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$
. So, the total time taken will be $ = 20\,s$
. Hence, the average speed will be $ = \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 $10\,m{s^{ - 1}}$
. 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.
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