Physics homework help please?

Question

A rubber ball with a mass of 200 g is released from rest from a height of 2.0 m. It falls to the floor, bounces, and rebounds. The graph at right depicts the magnitude of the upward normal force that the floor exerts on the ball at various instants in time. The graph only shows the narrow window of time surrounding the interval when the ball was in contact with the floor. e.) What is the impulse on the ball by the earth’s gravitational pull during the same 10 ms? f.) By how much does the ball’s momentum change as a result of this 10-ms period? g.) How high does the ball rebound?

There's also a graph: http://i179.photobucket.com/albums/w298/simplestarlight/49794.jpg

Answer 1

(e) Gravity always act downwards. It is present during the 10ms interval while the ball is in contact with the floor. To find impulse, you multiply force with time:

Impulse_gravity=10ms*F_gravity
=0.010sec*0.2Kg*9.8m/sec^2
=0.0196 Kg*m/sec

(f) Change in momentum is the sum of impulses, one from gravity and one from floor. You find the impulse from the floor as the area under the triangle in the graph.

Impulse_floor=0.5*0.010sec*480N
=2.4 Kg*m/sec
Delta_momentum=2.4-0.0196
=2.38 Kg*m/sec

(g) To find the height of the rebound, you first need to find the velocity at which the ball hits the floor. You use the equation Vf^2=Vi^2=2*a*s where Vi=0, a=9.8m/sec^2, and s=2m.

V_ball=sqrt(2*9.8*2)
=6.261 m/sec

The momentum of the ball when it hits the floor is just the velocity times mass:

P_ball_down=m*Vf
=1.251 Kg*m/sec downward

Now apply the change in momentum from part (f) to find the momentum of the ball right as it leaves the floor on the way back up:

P_ball_up=1.1282 Kg*m/sec upward

Divide through by mass again to find upward velocity:

V_ball_up=P_ball_up/m
=5.641 Kg*m/sec

Now use the equation of motion again to find distance s:

s=V_ball_up^2/(2*g)
=1.62m

Voila!

Answer 2

The time of fall is obtained from s = 0.5gt*2, or 2 = 0.5(9.8)t^2, so t = 0.6389 seconds. Momentum at impact will be the velocity times the mass; the velocity is 9.8(0.6389) = 6.2612 m/sec, so the momentum is 0.2(6.2612) = 1.2522 kg-m/sec. The impulse from the graph is 2.4 kg-m/sec so the momentum after the collision is 1.1478 kg-m/sec upward. (The effect of the earth's gravity during the collision is negligible.) To get the height of the rebound, divide out the mass and use s = 0.5v^2/g.