Forest UCM MnAM InElasticCol

An Inelastic collision conservers Momentum But Not energy

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

If there are no external force then

Conservation of momentum

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

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

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

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 [math]\displaystyle{ M }[/math]. Each hobo has a mass [math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ u }[/math] the railcar will move in the opposite direction at some speed [math]\displaystyle{ v }[/math]. Since the hobo's speed is relative to the car, the hobos speed relative to the ground is [math]\displaystyle{ u-v }[/math].

conservationof momentum

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

or

[math]\displaystyle{ 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]\displaystyle{ 0 = m(u-v_1) - (m+M) v_1 }[/math]

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

But now the second hobo jumps off

[math]\displaystyle{ m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2 }[/math]

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

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

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

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

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

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