Question

A blue puck with mass 0.0400 kg, sliding with a speed of 0.200 m/s on a frictionless, horizontal surface, makes a perfectly elastic head‐on collision with a red puck with mass m, initially at rest. After the collision, the velocity of the blue puck is 0.0500 m/s in the same direction as its initial velocity. (a) Find the mass m of the red puck, and (b)the velocity (magnitude and direction) of the red puck after the collision

Answer 1

Part(a): The velocity of the red puck is
`\bf{0.2500~m/s}` and the direction will be the same as initial direction of blue puck.

Part(b): The mass of the red puck is
`\bf{0.024~Kg}`.

Step-by-step explanation:

Given:

The mass of the blue puck,
`m_(b) = 0.0400~Kg`

The initial velocity of the blue puck,
`v_(ib) = 0.200~m/s`

The final velocity of the blue puck,
`v_(fb) = 0.0500~m/s`

The initial velocity of the red puck,
`v_(ir) = 0~m/s`

Consider the mass of the red puck be
`m_(r)` and its final velocity be
`v_(fr)`.

Part(a):

The relation between the relative velocities of the particles is given by

`v_(ib) - v_(ir) = v_(fr) - v_(fb)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)`

Substitute the values in equation (1)

`~~~~&& 0.200~m/s - 0 = v_(fr) - 0.0500~m/s\\&or,& v_(fr) = 0.2500~m/s`

Part(b):

From momentum conservation,

`m_(b)v_(ib) + m_(r)v_(ir) = m_(b)v_(fb) + m_(r)v_(fr)~~~~~~~~~~~~~~~~~~~~~~~~(2)`

Substitute the values in equation (2).

`~~~~(0.0400~Kg)(0.200~m/s) - 0 = m_(r)(0.2500~m/s) + (0.0400~Kg)(0.05~m/s)\\&or,& m_(r) = 0.024~Kg`