A block of mass m 270 kg slides down a 300 incline which is

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.

Solution

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