A block of mass m = 2.70 kg slides down a 30.0 incline which is 3.60 m high. At

Question

A block of mass m = 2.70 kg slides down a 30.0 incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.00 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 = 2.70 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 = 6.00 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.

Part B

Determine the speed of the block with mass M = 6.00 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 energy conservation to find the speed of block m as just before it hits block M.

m v^2 /2 = m gh

v =sqrt(2 x 9.8 x 3.60 ) = 8.4 m

suppose after collision block m goes back with speed v1 and block M goes ahead with v2 th2n

for elastic collision,

velocity of approach = velocity of seperation

v = v1 + v2

v2 = 8.4 - v1

now applying momentum conservation,

mv + M*0 = - mv1 + Mv2

2.7 x 8.4 = - 2.7v1 + 6v2

22.68 = -2.7v1 + 6(8.4 - v1)

27.72 = 8.7v1

v1 = 3.186 m/s .......Ans

(B) v2 = v - v1 = 8.4 - 3.19 = 5.214 m/s

(C) using energy conservation again,

m v1^2 /2 = m g h   and h = Lsin30

v1^2 = 2 g Lsin30

L = (3.186^2) / (2 x 9.8 x sin30) = 1.04 m .......Ans

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