A Girl on a Trampoline

Part A = 4.98m/s

Part B = 3.98m/s

Part C = inelastic

Part D = 2.81m

Solutions Below:

_{1}= 60 kilograms springs from a trampoline with an initial upward velocity of v

_{i}= 8.0 meters per second. At height h = 2.0 meters above the trampoline, the girl grabs a box of mass m

_{2}= 15 kilograms.

For this problem, use g = 9.8 meters per second per second for the magnitude of the acceleration due to gravity.

Part A

_{before}of the girl immediately before she grabs the box?

We can solve this using the kinematic equation V_{f}^{2} = V_{i}^{2} + 2a(d):

V_{f}^{2} = V_{i}^{2} + 2a(d)

V_{f}^{2} = (8.0)^{2} + 2(-9.8)(2.0)

V_{f}^{2} = 64 – 39.2

V_{f}^{2} = 24.8

V_{f} = 4.98m/s

Part B

_{after}of the girl immediately after she grabs the box?

Since momentum is conserved, we can solve this equation using the formula for momentum (p = mv). Note that since this is an inelastic collision, energy is not conserved, so we won’t be able to solve using the formulas for kinetic and potential energy:

p = mv

Solve for momentum using the velocity (from Part A) just before the girl picks up the box:

p = 60(4.98)

p = 298.8Ns

Now solve for when the box gets picked up:

p = mv

298.8 = (60 + 15)v

298.8 = 75v

v = 3.98m/s

Part C

A. elastic

B. inelastic

Since the two objects stick together after colliding, the collision is inelastic:

B. inelastic

Part D

_{max}that the girl (with box) reaches? Measure h

_{max}max with respect to the top of the trampoline.

We can solve this using the kinematic equation V_{f}^{2} = V_{i}^{2} + 2a(d). We know that at the maximum height, gravity will have slowed velocity to zero, and our initial velocity will be from just after the girl picks up the box (from Part B):

V_{f}^{2} = V_{i}^{2} + 2a(d)

0 = (3.98)^{2} + 2(-9.8)(d)

0 = 15.84 – 19.6d

15.84 = 19.6d

d = 0.808

Since the girl picked up the box at 2m above the trampoline, we need to add that into the height:

d = 0.808 + 2

d = 2.81m