Question3 18 pts Two rubber balls with two different masses are dropped from hei

Question

Question3 18 pts Two rubber balls with two different masses are dropped from height of 5 meters from rest (neglect air resistance and assume the rubber balls are small in size so they both start at 5 meters above the ground). The more massive ball of 1.5 kg is below the less massive ball of 0.4 kg. The 1.5 kg hits the ground first and collides with the Earth which is at rest) in a perfectly collision. Then, just above the ground (say at the ground), the 1.5 kg ball (after colliding with the Earth and changing its velocity) collides with smaller mass in another perfectly elastic collision. After the smaller mass collides with other rubber ball, to what maximum height above the ground does it reach in meters?

Explanation / Answer

From Conservation of Energy, the speed of each ball just before hitting the ground will be:

v = [2gh]1/2 = [2 x 9.8 x 5]1/2 = 9.9 m/s

Let upwards be positive direction of motion. Now, after the elastic collision of the bigger ball, the momentum of the system of two balls before collision with each other will be:

pi = 1.5(9.9) - 0.4(9.9) = 10.89 kgm/s

due to conservation of momentum, the momentum of the system will be the same after the collision with each other.

=> 10.89 = 1.5V + 0.4v

=> V = 7.26 - 0.2667v

also, the kinetic energy is conserved

so, (1/2)(1.5)(9.9)2 + (1/2)(0.4)(9.9)2 = (1/2)(1.5)V2 + (1/2)(0.4)v2

=> 186.219 = 1.5V2 + 0.4v2

substitute for V from above to get:

186.219 = 1.5[7.26 - 0.2667v]2 + 0.4v2

=> 186.219 = 1.5[52.7076 + 0.0711v2 - 3.8725] + 0.4v2

=> 186.219 = 79.0614 + 0.1066v2 - 5.8087 + 0.4v2

=> 0.5066v2 - 112.966 = 0

=> v = 14.93 m/s

this will be the velocity of the smaller ball after the collision.

now, use v2 = u2 - 2gH

at the highest point, v = 0 m/s

therefore,0 = 14.932 - 2(9.8)H

=> H = 11.377 m

this is the maximum height till which the smaller ball will rise.

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