Question

A 150-kg roller coaster car moving with a velocity of 4 m/s to the right hits a 125-kg car that is standing still. If the two cars hit and stick together, what is their final velocity?

Answer 1

To determine the velocity of the two cars that stuck together in a collision, we use the law of conservation of momentum: the total momentum of the system before and after the collision remains the same

$${{\vec{p}}_{1}}+{{\vec{p}}_{2}}=\text{constant}$$

Recall that the momentum of an object is the product of its mass m and velocity $\vec{v}$

$$\vec{p}=m\vec{v}$$

Then the law of conservation of momentum is written as

$${{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}={{m}_{1}}{{\vec{u}}_{1}}+{{m}_{2}}{{\vec{u}}_{2}}$$

where ${{m}_{1}}=150\,kg,\,\,{{m}_{2}}=125\,kg$ are the masses of the first and second cars, ${{\vec{v}}_{1}}=4\,m/s,\,{{\vec{v}}_{2}}=0\,m/s$ are their velocities before the collision and ${{\vec{u}}_{1}}={{\vec{u}}_{2}}=\vec{u}$ are their velocities after the collision (final velocity). Also define the direction of the initial velocity vector to be the x-direction. Now this equation becomes

$${{m}_{1}}{{v}_{1}}\cdot \vec{i}+0=\left( {{m}_{1}}+{{m}_{2}} \right)\vec{u}$$

Find the final speed

$$\vec{u}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{1}}+{{m}_{2}}}\vec{i}$$

Substituting the given numbers:

$$\vec{u}=\left( \frac{150\,kg\cdot 4\,m/s}{150\,kg+125\,kg} \right)\vec{i}\approx \left( 2.2\,m/s \right)\vec{i}$$