Question

A block of mass 1 kg is horizontally thrown with a velocity of 10 m/s on a stationary long palko
mass 2 kg whose surface has u = 0.5. Plank rests on frictionless surface. Find the time when block
comes to rest w.r.t. plank.
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Answer 1

Answer:

Explanation:

=> When block  comes to rest with respect to plank, the common velocity of the system is suppose 'V'

=> According to the law of conservation of momentum,

1 * 10 = (2 + 1)V

10 = 3V

V = 10/3

the force on block  = .5 * 1 * 10 = 5 N

deacceleration of block  a = 5/1 = 5 m/s²

=> suppose 't' time is required to change velocity of block from 10 to 10/3.

Thus, 10−10/3 = at =5t

5t = 30-10 / 3

5t = 20 / 3

t = 20/ 5*3

t = 4 / 3 sec.

Therefore, the time when block  comes to rest with respect to plank is 4/3 sec.

Answer 2

Answer :-

F = v dm/dt

= 5 x 1

= 5N

acceleration= a = F/m

= 5/2

= 2.5 ms⁻²