Question

Two hobos, each of mass mh , are standing at one end of a stationary railroad flatcar with frictionless wheels and mass m t-c . Either hobo can run to the other end of the flatcar and jump off with the same speed u (relative to the car). (a) Use conservation of momentum to find the speed of the recoiling car if the two men run and jump simultaneously. (b) What is it if the second man starts running only after the first has already jumped? Which procedure gives the greater speed to the car? [Hint: The speed u is the speed of either hobo, relative to the car just after he has jumped; it has the same value for either man and is the same in parts (a) and (b).]

Answer 1

Answer and Solution:

As per the question:

Mass of each hobo is $m_{h}$

Velocity of the either hobo is 'u'

Mass of the flatcar is $m_{fc}$

Now,

(a) Suppose $v_{r}$ be the recoil velocity, then

Relative speed of either hobo w.r.t ground just after their jump is $u - v_{r}$

Now, by the principle of conservation of momentum:

$2m_{h}(u - v_{r}) = m_{fc}v_{r}$

$v_{r} = \frac{2m_{h}}{m_{fc} + 2m_{h}}\times u$                 (1)

(b) Suppose the recoil velocity of the flatcar be $v_{rc}$ after the jump of the first hobo and  $v_{rc}$ after the jump of the second hobo.

Now, applying the of conservation of momentum, after the jump of the first hobo is the same as done in part (a), but when only one hobo jumps, the mass of the second hobo remains:

$m_{h(u - v_{rc})} = (m_{h} + m_{fc}}v_{rc})$

$v_{rc} = \frac{m_{h}}{m_{fc} + 2m_{h}}\times u$                    (2)

Now, by conservation of momentum for the second jump:

$m_{h}(u - v'_{rc}) - m_{fc}v'_{rc} = - (m_{fc} + m_{h})v_{rc}$

$v'_{rc} = \frac{m_{h}u + (m_{h} + m_{fc})}{m_{h} + m_{fc}}\times v_{rc}$  (3)

Now, from eqn (1), (2) and (3):

It is clear that $v'_{rc}$ > v_{rc}