Question

A 0.250-kg ice puck, moving east with a speed of 5.48 m/s, has a
head-on collision with a 0.900-kg puck initially at rest. Assume
that the collision is perfectly elastic.

What is the **speed** of the 0.900-kg puck after the
collision?

Answer #1

here,

the mass of ice puck , m1 = 0.25 kg

the initial velocity of ice puck , u1 = 5.48 m/s

mass of second puck , m2 = 0.9 kg

let the final velocity of each puck be v1 and v2

using conservation of momentum

m1 * u1 = m1 * v1 + m2 * v2

0.25 * 5.48 = 0.25 * v1 + 0.9 * v2 .....(1)

and

using conservation of kinetic energy

0.5 * m1 * u1^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

0.25 * 5.48^2 = 0.25 * v1^2 + 0.9 * v2^2 ....(2)

from (1) and (2)

v1 = - 3.1 m/s

v2 = 2.38 m/s

the speed of the 0.9 kg block is 2.38 m/s

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