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Definition of a Projectile

A mass m 1 ...

Question

A mass $m_{1}$ with initial speed $v_{0}$ in the positive $x$ -direction collides with a mass $m_{2} = 2 m_{1}$ which is initially at rest at the origin, as shown in figure. After the collision $m_{1}$ moves off with speed $v_{1} = v_{0} / 2$ in the negative $y$ - direction, and $m_{2}$ moves off with speed $v_{2}$ at angle $θ$ . If the energy change due the collision is $\frac{m {v_{0}}^{2}}{10 + x}$ . Find $x$ .

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Solution

By the law of conservation of linear momentum,
along x-axis ,
total momentum before collision=total momentum after collision
$m_{1} v_{0} + 0 = m_{1} \times 0 + m_{2} v_{2} cos θ$ ,
or $m_{1} v_{0} = 2 m_{1} v_{2} cos θ$ , (given $2 m_{2} = 2 m_{1}$ ) ,
or $v_{0} = 2 v_{2} cos θ$ .......eq1 ,
along y-axis ,
total momentum before collision=total momentum after collision
$0 = - m_{1} v_{1} + m_{2} v_{2} sin θ$ ,
or $0 = - m_{1} v_{0} / 2 + 2 m_{1} v_{2} sin θ$ , (given $v_{1} = v_{0} / 2$ ) ,
or $v_{0} / 4 = v_{2} sin θ$ .......eq2,
by squarring and adding eq1 and 2 , we get
$v_{2} = \frac{\sqrt{5}}{4} v_{0}$ ,
Now , change in kinetic energy will be ,
$Δ k =$ kinetic energy after collision-kinetic energy before collision ,
or $Δ k = 1 / 2 m_{1} v_{1}^{2} + 1 / 2 m_{2} v_{2}^{2} - 1 / 2 m_{1} v_{0}^{2} + 0$ ,
or $Δ k = 1 / 2 m_{1} (v_{0}^{2} / 4) + 1 / 2 \times 2 m_{1} (\sqrt{5} / 4 v_{0})^{2} - 1 / 2 m_{1} v_{0}^{2}$ ,
or $Δ k = \frac{m_{1} v_{0}^{2}}{16} = \frac{m_{1} v_{0}^{2}}{10 + 6}$

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