Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Two perfectly elastic balls of same mass m are moving with velocities u 1 and u 2 . They collide head on elastically n times. The kinetic energy of the system finally is :
A
B
( 
+ 
)
C
m ( 
+ 
)
D
mn ( 
+ 
)
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: Calculate the initial kinetic energy of the system:
K.E. = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 = \frac{m}{2} (u_1^2 + u_2^2)
Step 2: In an elastic collision, kinetic energy is conserved. Therefore, after n elastic collisions, the total kinetic energy will remain the same as before any collisions
Step 3: Thus, the final kinetic energy after n collisions will still be:
K.E. = \frac{m}{2} (u_1^2 + u_2^2)
Therefore, A.
K.E. = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 = \frac{m}{2} (u_1^2 + u_2^2)
Step 2: In an elastic collision, kinetic energy is conserved. Therefore, after n elastic collisions, the total kinetic energy will remain the same as before any collisions
Step 3: Thus, the final kinetic energy after n collisions will still be:
K.E. = \frac{m}{2} (u_1^2 + u_2^2)
Therefore, A.
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