A 0.500 kg metal ball impact a 20.0 kg metal block resting on a horizontal surfa

Question

A 0.500 kg metal ball impact a 20.0 kg metal block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. After the impact , the ball rebounds with a speed equal to 0.200 m/sand produces a displacement of the block of 0.50 m measured from the position of the impact.

a) what was the speed of the block just after the ball impact?

b) what was the initial speed of the ball?

c)calculate the kinetic energy of the ball and block before and after the impact?

d)what kind of collision took place during the impact? explain

e) calculate the impulse of the block from the moment just after the impactto his final position.

Explanation / Answer

a) deceleration of block on rough surface = u g = 0.20 x 9.81 = 1.96 m/s^2

block comes to rest in 0.50 m.

using vf^2 - vi^2 = 2ad

0^2 - v^2 = 2(-1.96)(0.50)

v = 1.40 m/s

b) now using momentum conservation,

initial momentum of block ball system = final momentum

0.50v + 20*0 = 20x 1.40 - 0.5*0.2

v = 56.03 m/s

c) Before the impact :

KEi = 0.50 x 56.03^2 /2 = 784.8 J

after the impact:

KEf = 20*1.40^2 /2 + 0.5*0.2^2 /2

= 19.61 J

d) e = velocity of seperation / velocity of approach

e = (1.40+0.2) / 56.03 =0.03

this is an inelastic collision.

e) impulse = change in momentum = m * deltav

= 20 ( 1.40 - 0 ) = 28 kg m/s

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