Question

Two balls with the same mass are dropped simultaneously from the same height above the floor. The first bounces all the way up to its original height (perfectly elastic), while the second sticks to the floor on impact (perfectly inelastic). Which ball delivers the greater impulse to the floor?

A. the first.
B. the second.
C. neither since they're equal.
A ball of mass m 50 g is dropped from a height of h 0.92 m above the floor, and then bounces all the way up to its original height (perfectly elastic). Calculate the impulse delivered by the ball to the floor. ______________kg m/s (0.02 kg m/s)

Answer 1

1) A

2) 0.42 kg.m/s

Step-by-step explanation:

Here we need to apply the Impulse-Momentum theorem:

`J=\Delta p\\J=m*\Delta v\\`

For the first ball because it bounces to its original height the velocity just before the impact is the same right after but in the opposite direction, so:

`J=m*(v_f-v_i)\\J=m*(-v_i-v_i)\\J=-2*m*vi`

because the velocity just before the impact is going down the velocity is negative so:

J=2*m*vi

for the second ball, the velocity right after the impact is zero, so:

`J=m*(0-v_i)\\J=m*(0-v_i)\\J=-m*vi`

J=m*vi

As you can see the first ball has a greater impulse.

for the second problem, we need to apply the last formula:

`J=m*(v_f-v_i)`

`v_f^2=v_o^2+2*a*\Delta y\\v_f=√(2*9.8*0.92m)\\v_f=4.2m/s`

that is 4.2m/s going down or -4.2m/s.

this final velocity is the velocity just before the colition.

`J=50*10^(-3)kg*(4.2m/s-(-4.2m/s))\\J=0.42kg.m/s`