Question

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the velocity of car A after the collision?

Answer 1

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;

$v_{A1}$ = the initial velocity of car A;

$v_{A2}$ = the final velocity of car A;

and

$m_B$ = mass of car B;

$v_{B1}$ = the initial velocity of car B;

$v_{B2}$ = the final velocity of car B.

Then, the law of conservation of momentum demands that

$m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}$

And the conservation of kinetic energy says that

$\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2$

These two equations are solved for final velocities  $v_{A2}$ and $v_{B2}$ to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

$\boxed{v_{A2} = -1.05m/s}$

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

$\boxed{v_{B2} = 3.49m/s}$

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.