Forest UCM MnAM InElasticCol

An Inelastic collision conservers Momentum But Not energy

Consider a collision between two bodies of mass $m_1$ and $m_2$ initially moving at speeds $v_1$ and $v_2$ respectively. They stick together after they collide so that each is moving at the same velocity $v$ after the collision.

If there are no external force then

Conservation of momentum

$m_1 v_1 + m_2 v_2 = \left (m_1 + m_2 \right ) v$

Given that the amsses and initially velocities are known we can solve for v such that

$v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$

Forest_UCM_MnAM#Inelastic_Collision_of_2_bodies

problem 3.4 two hobos

Two hobos are standing at one end of a stationary railroad flatcar of mass $M$. Each hobo has a mass $m$ and the flatcar has frictionless wheels. A hobo can run to the other end of the car and jump off with a speed $u$ with respect to the car.

a.) Find the final speed of the car if both men run and jump off simultaneously.

At the instant the hobo jumps off with speed $u$ the railcar will move in the opposite direction at some speed $v$. Since the hobo's speed is relative to the car, the hobos speed relative to the ground is $u-v$.

conservationof momentum

$\Rightarrow 0 = 2m(u-v) - Mv$

or

$v = \frac{2m}{M+2m}u$

b.) Now consider the case where they jump separately. The second hobo starts running AFTER the first hobo has jumped off.

Break the problem up into the two parts

PART A: The first hobo jumps off

Conservation of momentum

$0 = m(u-v_1) - (m+M) v_1$

$v_1 = \frac{m}{M+2m}u$: same as before

But now the second hobo jumps off

$m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2$

$-(m+M)\frac{m}{M+2m}u= mu- (M+m) v_2$

$(M+m) v_2= mu + (m+M)\frac{m}{M+2m}u$

$(M+m) v_2= \frac{m(M+2m)+m(m+M)}{M+2m}u$

$(M+m) v_2= \frac{m(M+2m+m+M)}{M+2m}u$

$(M+m) v_2= \frac{m(2M+3m)}{M+2m}u$

$v_2= \frac{m(2M+3m)}{(M+2m)(M+m)}u$