Question

On a safety test course, a 1000 kg car heading north at 5 m/s collides head-on with an 800 kg car heading south at 4 m/s. At these low speeds, the new high-tech bumpers prevent the cars from crumpling; they bounce off each other. After the bounce, the 1000 kg car is heading southward at 1 m/s. What is the post-collision speed and direction of motion of the other car

Answer 1

Answer:

The speed is $3.5\frac{m}{s}$ and the direction is heading north.

Explanation:

In collisions the force exerted by the objects that collide is higher enough than the external forces that we can neglect that external forces, with that assumption we can use the conservation fo momentum law that states, final total momentum (pf) is equal initial total momentum (pi) if there’re not external forces or they are small enough to be neglected. Mathematically:

$p_f=p_i$

The total momentum is the sum of the momentum of each of the bodies we're dealing, in our case the moment of each car, then:

$p_{nf}+p_{sf}=p_{ni}+p_{si}$

with pn the momentum of the 1000kg car heading north and ps the 800kg car heading south. Momentum is defined as mass times velocity, then:

$m_nv_{nf}+m_sv_{sf}=m_nv_{ni}+m_sv_{si}$ (1)

It's important to note that when we talk about momentum and velocity direction matters, so we're are going to choose a system of reference where quantities pointing north are positive and pointing south are negative. So, the initial velocity of 1000 kg car is vni=5 m/s, initial velocity of 800 kg car is vsi=-4 m/s and the final velocity of 1000 kg car is vnf=-1 m/s. Now we can solve (1) for vsf and use the values we already have:

$v_{sf}=\frac{m_nv_{ni}+m_sv_{si}-m_nv_{nf}}{m_s}=\frac{(1000)(5)+(800)(-4)-(1000)(-1)}{800}$

$v_{sf}=3.5\frac{m}{s}$

Because the sign is positive the direction is to heading north.