Question

A 2190 kg car moving east at 10.7 m/s collides with a 3210 kg car moving east. The cars stick together and move east as a unit after the collision at a velocity of 5.46 m/s. a) What is the velocity of the 3210 kg car before the collision?

Answer 1

Given data

*The given mass of the car is m = 2190 kg

*The car is moving east at a speed is u = 10.7 m/s

*The mass of the second car is M = 3210 kg

*Both the cars moving with a velocity is v = 5.46 m/s

The expression for the velocity of the 3210 kg car before the collision is given by the conservation of momentum as

$\begin{gathered} P_i=P_f \\ mu+MV=(m+M)v \end{gathered}$

*Here V is the velocity of the 3210 kg car before the collision

Substitute the known values in the above expression as

$\begin{gathered} (2190)(10.7)+(3210)V=(2190+3210)\times(5.46) \\ 23433+3210V=29484 \\ 3210V=29484-23433 \\ V=\frac{6051}{3210} \\ V=1.88\text{ m/s} \end{gathered}$

Hence, the velocity of the 3210 kg car before the collision is V = 1.88 m/s