In Fig. 9-69 puck 1 of massis sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a distancefrom the base of the bench. What is the mass of puck 2? (Hint: Be careful with signs)

The mass of puck 1 is, $m_1=0.20\mathrm{kg}$.

The displacement $∆x_{2}is,∆x_2=d$.

The displacement $∆x_1\mathrm{is},∆{\mathrm{x}}_1=-2\mathrm{d}$.

By using the equation for the velocities${\mathit{v}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{v}}_{\mathbf{2}}$for elastic collision, find the equation for displacementsrole="math" localid="1661314976098" $\mathbf{∆}{\mathit{x}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}{\mathbf{x}}_{\mathbf{2}}$. Finally, dividing$\mathbf{∆}{\mathit{x}}_{\mathbf{1}}\mathbf{}\mathbf{by}\mathbf{}\mathbf{∆}{\mathbf{x}}_{\mathbf{2}}$and putting all given values, find the massof puck 2.

Formulae are as follow:

The displacement $∆x\mathrm{is},∆\mathrm{x}=\mathrm{vt}$,

The velocity role="math" localid="1661315182285" ${\mathrm{v}}_2\mathrm{is},{\mathrm{v}}_2=\frac{2m_1}{m_1+m_2}{v}_{2i}$,

The velocity${\mathrm{v}}_1\mathrm{is},{\mathrm{v}}_1=\frac{m_1-m_2}{m_1+m_2}{v}_{1i}$

Where, $∆x$is change in displacement, v is velocity, m is mass andt is time.

Note that both pucks have the same time of fall t. Thus,

$∆x_2=v_{2}t$

Where, $$\begin{gathered} ∆x_2=d\mathrm{and}{\mathrm{v}}_2=\frac{2{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}{\mathrm{v}}_{1\mathrm{i}} \\ ∆{\mathrm{x}}_1={\mathrm{v}}_1\mathrm{t} \\ \mathrm{Where},∆{\mathrm{x}}_1=-2\mathrm{d}\mathrm{and}{\mathrm{v}}_1=\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}{\mathrm{v}}_{1\mathrm{i}} \end{gathered}$$

Dividing the first equation by the second,

$$\begin{gathered} \frac{∆x_2}{∆x_1}=\frac{v_{2}t}{v_{1}t} \\ \frac{d}{-2d}=\frac{\left({\displaystyle \frac{2m_1}{m_1+m_2}}{v}_{1i}\right)}{\left({\displaystyle \frac{m_1-m_2}{m_1+m_2}}{v}_{1i}\right)} \\ \frac{1}{\left(-2\right)}=\frac{\left({\displaystyle \frac{2m_1}{m_1+m_2}}\right)}{\left({\displaystyle \frac{m_1-m_2}{m_1+m_2}}\right)} \\ \left(\frac{m_1-m_2}{m_1+m_2}\right)=\left(-2\right)\left(\frac{2m_1}{m_1+m_2}\right) \\ {m}_1-m_2=-4m_1 \\ {m}_2=4m_1+m_1 \\ {m}_2=5m_1 \\ {m}_2=5\times 0.2 \\ {m}_2=1.0\mathrm{kg} \end{gathered}$$

Hence, the mass of puck 2 is $m_2=1.0\mathrm{kg}$.

Therefore, by using the equation for the velocities ${\mathrm{v}}_1\mathrm{and}{\mathrm{v}}_2$for elastic collision, the mass can be found.