Question

A 1.40-kg ball, moving to the right at a velocity of +2.87 m/s on a frictionless table, collides head-on with a stationary 6.70-kg ball. Find the final velocities of (a) the 1.40-kg ball and of (b) the 6.70-kg ball if the collision is elastic. (c) Find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.

Answer #1

Part A

In a perfectly elastic collision

Using momentum conservation

Pi = Pf

m1V1i + m2V2i = m1V1f + m2*V2f

given that m1 = 1.40 kg & m2 = 6.70 kg

V1i = +2.87 m/sec

V2i = 0

1.40*2.87 + 6.70*0 = 1.40*V1f + 6.70*V2f = 4.018

Now other condition of the elastic collisions is that

V1f - V2f = V2i - V1i

V1f - V2f = 0 - 2.87

**V1f - V2f = -2.87**

**1.40*V1f + 6.70*V2f = 4.018**

Now Solving both equation

Multiply equation 1 with 6.70 and Add both of them

8.10*V1f = (-2.87*6.70 + 4.018)

V1f = (-2.87*6.70 + 4.018)/8.10 = -1.878 m/sec

**V1f = -1.878 m/sec (-ve sign means ball is reflected
backward)**

V2f = V1f + 2.87

**V2f = -1.878 + 2.87 = 0.992 m/sec**

Part C

when Collision is inelastic

Pi = Pf

m1v1 + m2v2 = (m1 + m2)*V

V = (1.40*2.87 + 6.70*0)/(1.40 + 6.70)

V = 0.496 m/sec (direction will towards right)

Please Upvote.

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