Question

1)A 861 g rubber ball is dropped from height 15.1 m, and it rebounds to height 3.15 m. Assuming there is no air friction, find the magnitude of the impulse, in N-s, exerted by the ground on the ball.

2)Milly (mass 98.7 kg) runs at 2.73 m/s along the positive x-axis and then jumps onto a stationary 278 kg box. Milly and the box slide across the floor for distance 0.357 m, and then they come to a stop. Find the coefficient of kinetic friction between the box and the floor.

3)Two astronauts float in deep space, at rest relative to each other. Including the equipment they carry, the first astronaut has total mass 521 kg and the second astronaut has total mass 615 kg. The first astronaut throws a 656 kg box of tools at 4.66, and the second astronaut catches the box. After the throw and catch: find the speed of one astronaut relative to the other, in m/s.

Answer 1

Multiple questions posted , so , answering only the first question:

Potential energy=mgh where m is mass , g is gravitational acceleration, h is height,

kinetic energy=1/2mv², where m is mass and v is velocity.

Consider the downward motion: Initial height=15.1 metres. So, initially, potential energy=m*9.8*15.1=147.98 m, where m is mass.

Also,final kinetic energy=1/2mv², where v is final velocity just before impact

Energy is conserved. So, final energy=initial energy => 1/2mv²=147.98 m => 1/2v²=147.98 => v²=2*147.98=295.96

=> v=17.2 m/s

Now,consider the upward motion: Final height=3.15 meters. So, finally, potential energy=m*9.8*3.15=30.87 m, where m is mass.

Also,initial kinetic energy=1/2mu², where u is the initial upward velocity after impact.

Energy is conserved So, 1/2mu²=30.87 m => 1/2u²=30.87 => u²=2830.87=61.74 => u=7.857 m/s

Given mass=861 g=0.861 kg

So, impulse transferred=m(v+u)=0.861*(17.2+7.857)=21.57 N-s