Question

1.On a frictionless, horizontal air table, puck A with a mass of 0.25 kg is moving to the right towardpuck B with a mass of 0.35 kg, which is initially at rest. After the collision, puck A has a velocity of0.12 m/s to the left, and puck B has a velocity of 0.65 m/s to the right. What was the speed of puckA before the collision?0.95 m/s2.5 m/s1.2 m/s0.79 m/s

Answer 1

Given data:

Mass of puck A:

$m_A=0.25\text{ kg}$

Mass of puck B:

$m_B=0.35\text{ kg}$

Initial velocity of puck B:

$u_B=0$

as the puck B is initally at rest.

Final velocity of puck A:

$v_A=-0.12\text{ m/s}$

Here, negative sign indicates that the velocity of the puck A is towards left.

Final velocity of puck B:

$v_B=0.65\text{ m/s}$

According to the conservation momentum, the momentum before and after collision remains constant that is,

$m_Au_A+m_Bu_B=m_Av_A+m_Bv_B$

Here, uA is the velocity of puck A before collision.

Rearranging the above equation in order to get an expression uA,

$u_A=\frac{m_Av_A+m_Bv_B-m_Bu_B}{m_A}$

Susbtituting all known values,

$\begin{gathered} u_A=\frac{(0.25\text{ kg})\times(-0.12\text{ m/s})+(0.35\text{ kg})\times(0.65\text{ m/s})-(0.35\operatorname{kg})\times(0)}{(0.25\text{ kg})} \\ =0.79\text{ m/s} \end{gathered}$

Therefore, the velocity of puck A before collision is 0.79 m/s.