Question

A 1090 kg car moving +11.0 m/s

Answer 1

The mass of the second car is 1434.21 kg

Explanation:

Using law of conservation of momentum,

          $m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given:

$m_{1}$ = 1090 kg

$u_{1}$ = 11 m/s

$u_{2}$ = 0

v = 4.75 m/s

We need to find $m_{2}$

When substituting the given values in the above equation, we get

$(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75$

$11990=5177.5+4.75 m_{2}$

$4.75 m_{2}=11990-5177.5$

$4.75 m_{2}=6812.5$

$m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}$