Question

A 78kg car traveling at 60m/s collides head on with 62 kg car traveling at 47 m/s causing them to stick together. They continue to move until they collide a 35-degree angle with a fire hydrant at causing the 2kg mass cap to fly off .05 seconds later at a 60-degree angle relative to the Y-axisWhat is the momentum of the cars before collision?What is the final velocity of the 62kg car?

Answer 1

Answer:

a). 7594 kg m/s^2

Explanation:

The initial momentum of the cars before the collision is the sum of the individual momenta

The momentum of the first car is

$p_1=m_1v_1=78\operatorname{kg}\cdot60m/s=4680\operatorname{kg}m/s$

The momentum of the second car is

$p_2=m_2v_2=62\operatorname{kg}\cdot47m/s=2914\operatorname{kg}\cdot m/s$

Hence, the total momentum is

$p_{\text{tot}}=p_1+p_2=4680+2914=\boxed{7594\operatorname{kg}\cdot m/s}$

which is our answer!

When the cars collide, the conservation of momentum gives

$p_1+p_2=(m_1+m_2)v_f$

where vf is the final velocity of the stuck-together cars.

solving for vf gives

$v_f=\frac{p_1+p_2}{m_1+m_2}$

since p1 + p2 = 7594 and m1 = 78 kg and m2 = 62kg; therefore,

$v_f=\frac{7594}{62+78}$$\boxed{v_f=54.24m/s}$

part (ii).

A sketch of the situation is given below.