Question

A car weighs 3600 kg is traveling at 21. 0 m/s. The driver doesn?t notice a red light and rear-ends another car at rest. Both cars stick together and move with a speed of 9. 0 m/s. What is the mass of the car that was at rest?.

Answer 1

Answer:

$4800\; {\rm kg}$.

Explanation:

When an object of mass $m$ travels at a velocity of $v$, the momentum $p$ of that object would be $p = m\, v$.

Momentum is conserved immediately after the collision. Hence, the momentum that the $3600\; {\rm kg}$ vehicle lost in the collision would be equal to the momentum that the other vehicle has gained.

Momentum that the $3600\; {\rm kg}$ vehicle lost:

$\begin{aligned}(\text{momentum change}) &= (\text{new momentum}) - (\text{original momentum}) \\ &= m\, v(\text{new}) - m\, v(\text{old}) \\ &= m\, (v(\text{new}) - v(\text{old})) \\ &= 3600\; {\rm kg} \, (9.0\; {\rm m\cdot s^{-1}} - 21.0\; {\rm m\cdot s^{-1}}) \\ &= (-43200)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}$.

The vehicle that was originally at rest would have gained that $43200\; {\rm kg \cdot m\cdot s^{-1}}$ of momentum. The momentum of that vehicle was $0\; {\rm kg \cdot m \cdot s^{-1}}$ before the collision since that vehicle was initially not moving. After gaining the $43200\; {\rm kg \cdot m\cdot s^{-1}}\!$ of momentum, the new momentum of that vehicle would be $p = 43200\; {\rm kg \cdot m\cdot s^{-1}}\!\!$.

Rearrange the equation $p = m\, v$ to obtain an expression for mass: $m = (p / v)$. It is given that the velocity of this vehicle is $v = 9.0\; {\rm m\cdot s^{-1}}$ after the collision. Substitute both momentum $p$ and velocity $v$ into the equation $m = (p / v)\!$ to find mass $m\!$:

$\begin{aligned} p &= \frac{m}{v} \\ &= \frac{43200\; {\rm kg \cdot m \cdot s^{-1}}}{9.0\; {\rm m\cdot s^{-1}}} \\ &= 4800\; {\rm kg} \end{aligned}$.

In other words, the mass of the vehicle that was originally not moving would be $4800\; {\rm kg}$.