Forest UCM MnAM InElasticCol

An Inelastic collision conserves Momentum But Not energy

[math]\vec{P}_{\mbox{initial}} = \vec{P}_{\mbox{final}}[/math]

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

If there are no external force then

Conservation of momentum

[math]m_1 v_1 + m_2 v_2 = \left (m_1 + m_2 \right ) v[/math]

Given that the masses and initially velocities are known we can solve for [math]v[/math] such that

[math]v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}[/math]

problem 3.4 two hobos

Two hobos are standing at one end of a stationary railroad flatcar of mass [math]M[/math]. Each hobo has a mass [math]m[/math] and the flatcar has frictionless wheels. A hobo can run to the other end of the car and jump off with a speed [math]u[/math] 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 [math]u[/math] the railcar will move in the opposite direction at some speed [math]v[/math]. Since the hobo's speed is relative to the car, the hobos speed relative to the ground is [math]u-v[/math].

conservationof momentum

[math]\Rightarrow 0 = 2m(u-v) - Mv[/math]

or

[math]v = \frac{2m}{M+2m}u[/math]

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

[math]0 = m(u-v_1) - (m+M) v_1[/math]

[math]v_1 = \frac{m}{M+2m}u[/math]: same as before

But now the second hobo jumps off

[math]m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2[/math]

[math]-(m+M)\frac{m}{M+2m}u= mu- (M+m) v_2[/math]

[math](M+m) v_2= mu + (m+M)\frac{m}{M+2m}u[/math]

[math](M+m) v_2= \frac{m(M+2m)+m(m+M)}{M+2m}u[/math]

[math](M+m) v_2= \frac{m(M+2m+m+M)}{M+2m}u[/math]

[math](M+m) v_2= \frac{m(2M+3m)}{M+2m}u[/math]

[math] v_2= \frac{m(2M+3m)}{(M+2m)(M+m)}u[/math]

if written in terms of the velocity from part (a)

[math] v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u =\frac{2M+3m}{2M+2m}v[/math]

The flatcar will be moving faster if the hobos jump off separately.

Forest_UCM_MnAM#Inelastic_Collision_of_2_bodies