Question

A block of mass m = 2.30 kg slides down a 30.0∘ incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.20 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.

a) Determine the speed of the block with mass m = 2.30 kg after the collision.

b)Determine the speed of the block with mass M = 6.20 kg after the collision.

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

Answer 1

First find speed of smaller mass before collision

when the smaller mass slides down the incline its potential energy is converted to kinetic energy.

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During collision momentum is conserved

Momentum before collision =Momentum after collision

-----------------(1)

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Since the collision is elastic kinetic energy is conserved

Kinetic energy before collision =Kinetic Energy after collision

Put (1) in above equation

Solve the quadratic equation using a calculator

(a)ANSWER:

-ve sign denotes that smaller block is going in opposite direction after collision, Opposite in direction with larger mass block.

(b)Put value of v_s in (1)

(b)ANSWER:

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

While going up kinetic energy is converted to potential energy

Consider the incline

ANSWER:

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