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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

User Tream
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1 Answer

4 votes

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=(m_Av_A+m_Bv_B-m_Bu_B)/(m_A)

Susbtituting all known values,


\begin{gathered} u_A=\frac{(0.25\text{ kg})*(-0.12\text{ m/s})+(0.35\text{ kg})*(0.65\text{ m/s})-(0.35\operatorname{kg})*(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.

User Alchi
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