Question

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

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

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 $V_{Af}\text{and}{V}_{Bf}$ . Use the equations from above. Let’s start with $V_{Af}$ .
$V_{Af}=\frac{m_A-m_B}{m_A+m_B}{V}_{Af}+\frac{(m_B-m_A)}{(m_A-m_B)}{V}_{Bi}$
Plug in our known values.
$V_{Af}=\frac{(10-2)kg}{(10+2)kg}.(2m/s)+\frac{2(2kg)}{(10+2kg)}.(-4m/s)$
$V_{Af}=\frac{8}{12}.(2m/s)+\frac{4}{12}.(-4m/s)$
$V_{Af}=\frac{16}{12}m/s+\frac{-16}{12}m/s$
$V_{Af}=0m/s$
The final velocity of the larger mass is zero. The collision completely stopped this mass.
Now for $V_{Bf}$
$V_{Af}=\frac{2m_A}{(m_A+m_B)}{V}_{Af}+\frac{(m_B-m_A)}{(m_A-m_B)}{V}_{Bi}$
Plug in our known values
$V_{Bf}=\frac{2(10kg)}{(10+2)kg}.2m/s+\frac{(2-10)kg}{(10+2kg)}.-4m/s$
$V_{Bf}=\frac{40}{12}m/s+\frac{32}{12}m/s$
$V_{Bf}=\frac{72kg}{12kg}m/s$
$V_{Bf}=6m/s$