Question

A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the collision. What is the velocity of the truck?

Answer 1

Answer:

The velocity of the truck is 3.33 m/s

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If some collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

There are two objects: The m1=4000 Kg car and the m2=6000 Kg truck. The car was moving initially at v1=4 m/s and the truck was at rest v2=0. After the collision, the car moves at v1'=-1 m/s. We need to find the velocity of the truck v2'. Solving for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

Substituting:

$\displaystyle v'_2=\frac{4000*4+6000*0-4000(-1)}{6000}$

$\displaystyle v'_2=\frac{16000+4000}{6000}$

$\displaystyle v'_2=3.33$

The velocity of the truck is 3.33 m/s