Question

# A block of mass m = 2.10 kg slides down a 30.0∘ incline which is 3.60...

A block of mass m = 2.10 kg slides down a 30.0∘ incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 8.00 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. Determine the speed of the block with mass m = 2.10 kg after the collision. Determine the speed of the block with mass M = 8.00 kg after the collision. Determine how far back up the incline the smaller mass will go.

Potential energy, PE = mgh = 2.1 x 9.8 x 3.6 = 74.09 J

Using law of conservation of energy

0.5 mu^2 = 74.09

Velocity u = sqrt(74.09 x 2/2.1) = 8.4 m/s

From law of conservation of momentum

For lighter block,

V1 = [(m1 - m2)/(m1 + m2)] u

Substituting values, we get

V1 = - 4.91 m/s (negative sign shows rebouncing)

For heavier block,

V2 = [2 m1/(m1 + m2)] u = 3.49 m/s

B)

From law of conservation of energy

m g h= 0.5 mv^2

9.8 x h = 0.5 x (-4.91)^2

Height, h = 1.23 m

Block will travel, 1.23/sin30 = 2.26 m