A block of mass m = 2.00 kg slides down a 30.0 incline which is 3.60 m high. At

Question

A block of mass m = 2.00 kg slides down a 30.0 incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.20 kg which is at rest on a horizontal surface (Figure 1) . (Assume a smooth transition at the bottom of the incline.) The collision is elastic, and friction can be ignored.

Part A

Determine the speed of the block with mass m = 2.00 kg after the collision.

Express your answer to three significant figures and include the appropriate units.

Part B

Determine the speed of the block with mass M = 7.20 kg after the collision.

Express your answer to three significant figures and include the appropriate units.

Part C

Determine how far back up the incline the smaller mass will go.

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Using Conservation of energy,m
calculate the velocity before collision of m
m*g*h =0.5 m*vi ^2
g*h =0.5 vi ^2
9.8 * 3.60 = 0.5*vi^2
vi = 8.4 m/s

for elastic collision,
e= -1
(Vf-vf) /(Vi-vi)= -1
(Vf-vf) / (0-0.84)= -1
Vf-vf = 0.84
Vf = 0.84+vf

Let the velocity after collision for m be vf and M be Vf
Use conservation of momentum during collision
m * vi =m * vf + M*Vf
2 * 8.4 = 2* vf + 7.20 * Vf
16.8 = 2* vf + 7.20 * Vf
16.8 = 2 *vf + 7.20 (0.84 + vf)
16.8 = 2 * vf + 6.048 + 7.20vf
10.752 = 9.2 vf
vf = 1.168 m/s
Vf = 0.84+vf
      = 0.84 + 1.168
      = 2.008 m/s
A)
Answer : vf =1.17 m/s
B)
Answer: Vf = 2.01 m/s
C)
For m, use conservation of energy
0.5 m * vf^2 = m*g*h
0.5 vf^2 = g*h
0.5 * 1.17^2 = 9.8 *h
h = 0.0698 m
Answer :
h = 0.0698 m

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