Question

A 1055-kg van, stopped at a traffic light, is hit directly in the rear by a 715-kg car traveling with a velocity of +2.25 m/s. Assume the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of (a) the car (b)the van?

Answer 1

Answer:

1.81779 m/s in the same direction as the car

0.43220 m/s in the same direction

Explanation:

$m_1$ = Mass of van= 1055 kg

$m_2$ = Mass of car = 715 kg

$u_1$ = Initial Velocity of van = 0 m/s

$u_2$ = Initial Velocity of car = 2.25 m/s

$v_1$ = Final Velocity of van

$v_2$ = Final Velocity of car

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\dfrac{m_1-m_2}{m_1+m_2}u_{1}+\dfrac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\dfrac{1055-715}{1055+715}\times 0+\dfrac{2\times 715}{1055+715}\times 2.25\\\Rightarrow v_1=1.81779\ m/s$

The final velocity of the van is 1.81779 m/s in the same direction as the car

$v_{2}=\dfrac{2m_1}{m_1+m_2}u_{1}+\dfrac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\dfrac{2\times 1055}{1055+715}\times 0+\dfrac{1055-715}{1055+715}\times 2.25\\\Rightarrow v_2=0.43220\ m/s$

The final velocity of the car is 0.43220 m/s in the same direction

Answer 2

Answer:

(a). The final velocity of the car is -0.432 m/s.

(b). The final velocity of the van is 1.82 m/s.

Explanation:

Given that,

Mass of van = 1055 kg

Mass of car = 715 kg

Initial velocity of car= 2.25 m/s

(a). We need to calculate the final velocity of the car

Using formula of velocity

$v_{c}=\dfrac{m_{v}-m_{c}}{m_{v}+m_{c}}\times u_{c}$

Put the value into the formula

$v_{c}=\dfrac{715-1055}{715+1055}\times2.25$

$v_{c}=-0.432\ m/s$

(b). We need to calculate the final velocity of the van

Using formula of velocity

$v_{v}=\dfrac{2m_{v}}{m_{v}+m_{c}}\times u_{c}$

Put the value into the formula

$v_{v}=\dfrac{2\times715}{715+1055}\times2.25$

$v_{v}=1.82\ m/s$

Hence, (a). The final velocity of the car is -0.432 m/s.

(b). The final velocity of the van is 1.82 m/s.