Question

A $10 \, kg$ mass travelling $2 \, m/s$ meets and collides elastically with a $2 \, kg$ mass travelling $4 \, m/s$ in the opposite direction. Find the final velocities of both objects.

• A. $V_{Af}$ = $2m/s$ ,​ $V_{Bf}$ = $3m/s$
• B. $V_{Af}$ = $0 \, m/s$ ,​ $V_{Bf}$ = $6 \, m/s$
• C. $V_{Af}$ = $5 \, m/s$ ,​ $V_{Bf}$ = $8 \, m/s$
• D. $V_{Af}$ = $4 \, m/s$ ,​ $V_{Bf}$ = $2 \, m/s$

Answer 1

Answer: (B)

We know the following
$m_{A}=10kg$ .
$V_{Ai}=2m/s$
$m_{B}=2kg$
$V_{Bi}=-4m/s$ . The negative sign is because the velocity is in the negative direction.
Now we need to find $V_{A f} \, \text{and} \, \, V_{B f}$ . Use the equations from above. Let’s start with $V_{A f}$ .
$V_{Af}=\frac{m_{A} - m_{B}}{m_{A} + m_{B}}V_{Af}+\frac{\left(\right. m_{B} - m_{A} \left.\right)}{\left(\right. m_{A} - m_{B} \left.\right)}V_{Bi}$
Plug in our known values.
$V_{Af}=\frac{\left(\right. 10 - 2 \left.\right) kg}{\left(\right. 10 + 2 \left.\right) kg}.\left(\right. 2 m / s \left.\right)+\frac{2 \left(\right. 2 kg \left.\right)}{\left(\right. 10 + 2 kg \left.\right)}.\left(\right. - 4 m / s \left.\right)$
$V_{Af}=\frac{8}{12}.\left(\right. 2 m / s \left.\right)+\frac{4}{12}.\left(\right. - 4 m / s \left.\right)$
$V_{Af}=\frac{16}{12}m/s+\frac{- 16}{12}m/s$
$V_{Af}=0m/s$
The final velocity of the larger mass is zero. The collision completely stopped this mass.
Now for $V_{B f}$
$V_{Af}=\frac{2 m_{A}}{\left(\right. m_{A} + m_{B} \left.\right)}V_{Af}+\frac{\left(\right. m_{B} - m_{A} \left.\right)}{\left(\right. m_{A} - m_{B} \left.\right)}V_{Bi}$
Plug in our known values
$V_{Bf}=\frac{2 \left(\right. 10 kg \left.\right)}{\left(\right. 10 + 2 \left.\right) kg}.2m/s+\frac{\left(\right. 2 - 10 \left.\right) kg}{\left(\right. 10 + 2 kg \left.\right)}.-4m/s$
$V_{Bf}=\frac{40}{12}m/s+\frac{32}{12}m/s$
$V_{Bf}=\frac{72 kg}{12 kg}m/s$
$V_{Bf}=6m/s$