Block 1, with a mass of m1 = 3 kg, slides on a frictionless surface and experien

Question

Block 1, with a mass of m1 = 3 kg, slides on a frictionless surface and experiences an elastic collision with Block 2, with a mass m2, which is initially at rest, then reverses direction at a quarter of its original speed, as shown below.

a. Express an equation for the relationship between the initial and final kinetic energy of m1 and m2.

b. Express an equation for the relationship between the initial and final momentum of m1 and m2.

c. Using the above equations, determine m2, i.e. the mass of Block 2.

Explanation / Answer

Here,

m1 = 3 Kg

a)
as the collision is elastic

intial kinetic energy = final kinetic energy

0.5 * m1 * V1o^2 = 0.5 * m1 * (V1o/4)^2 + 0.5 m2 * V2f^2

the equation is 0.5 * m1 * V1o^2 = 0.5 * m1 * (V1o/4)^2 + 0.5 m2 * V2f^2

b)

for the relation of the momentum m1 and m2

m1 * V1o = - m1 * V1o/4 + m2 * V2f

c)

for the equation 1

0.5 *3* V1o^2 = 0.5 * 3 * (V1o/4)^2 + 0.5 m2 * V2f^2 ---(3)

and

3 * V1o = -3 * V1o/4 + m2 * V2f

m2 * V2f = 13 * V1o/4

putting in 3

0.5 *3* V1o^2 = 0.5 * 3 * (V1o/4)^2 + 0.5 m2 * (13 * V1o/(4 * m2))^2

solving for m2

m2 = 3.76 Kg

the mass of m2 is 3.76 Kg

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