Question

A ball of mass m falls vertically from a height h and collides with a block of equal mass m moving horizontally with a velocity v on a surface. The coefficient of kinetic friction between the block and the surface is 0.2, while the coefficient of restitution (e) between the ball and the block is 0.5. There is no friction acting between the ball and the block. The velocity of the block decreases by

Answer 1

The velocity of the block decreases by v – v′ =  3√2gh

Explanation:

Using impulse = change in linear momentum along vertical axis on ball Ndt =  3/2 m√ 2gh

where as on block : μNdt = mv – mv′

=  2×3/2√2gh×m

= m (v – v′)

v – v′ =  3√2gh

Hence the velocity of the block decreases by v – v′ =  3√2gh

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A 100 N force applied on a block of mass 50 kg produces a constant velocity of 5 m/min on rough surface. The force friction exerted on the block will be?

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Answer 2

up one is correct but answer is 0.3 not 3 first