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Question: Two girls of equal mass \(m\)jump off the border line of a stationary carriage of mass \(M\)with the...

Two girls of equal mass mmjump off the border line of a stationary carriage of mass MMwith the same horizontal velocity uurelative to the carriage. Neglecting the effect of friction:

(A). They will impact greater velocity to the carriage by jumping off simultaneously.
(B). They will impact greater velocity to the carriage by jumping one after the other.
(C). They will impact greater velocity to the carriage in whatever manner they jump off.
(D). Insufficient data to reply.

Explanation

Solution

You can start by considering the case where the girls jump off simultaneously and use the concept of conservation of momentum i.e. 0=mu+mu−Mv0 = mu + mu - Mvto find the velocity of the carriage. Then use a similar method to calculate the velocity of carriage when girls jump off one after another. Then compare the velocities of the carriage in both the situations to reach the solution.

Complete step-by-step answer:
In the problem we are given a system with two girls of mass mm and a carriage of massMM. Now we have to figure out if the velocity of the carriage will be more when the two girls jump together or when they jump one after the other.
Case 1 – The girls jump off simultaneously.
Let the velocity of the carriage after the girls jump off bevv.
By using the concept of conservation of momentum, we get
0=mu+mu−Mv0 = mu + mu - Mv
Mv=2muMv = 2mu
v=2muMv = \dfrac{{2mu}}{M}(Equation 1)
Case 2 – The girl jumps off one after the other
Let the velocity of the carriage when one girl jumps off be v1{v_1}
By using the concept of conservation of momentum, we get
0=mu−(M+m)v10 = mu - (M + m){v_1}
mu=(M+m)v1mu = (M + m){v_1}
v1=muM+m{v_1} = \dfrac{{mu}}{{M + m}}
Let the velocity of the carriage when the second girls jumps off be v2{v_2}
By using the concept of conservation of momentum, we get
0=mu−Mv20 = mu - M{v_2}
mu=Mv2mu = M{v_2}
v2=muM{v_2} = \dfrac{{mu}}{M}
Let the net velocity of the carriage bev3{v_3}.
So, v3=v1+v2{v_3} = {v_1} + {v_2}
v3=muM+m+muM{v_3} = \dfrac{{mu}}{{M + m}} + \dfrac{{mu}}{M}
v3=mu(M+M+mM(M+m)){v_3} = mu\left( {\dfrac{{M + M + m}}{{M(M + m)}}} \right)
v3=muM(2M+mM+m){v_3} = \dfrac{{mu}}{M}\left( {\dfrac{{2M + m}}{{M + m}}} \right)
v3=2muM(M+m2M+m){v_3} = \dfrac{{2mu}}{M}\left( {\dfrac{{M + \dfrac{m}{2}}}{{M + m}}} \right)
v3=v(M+m2M+m){v_3} = v\left( {\dfrac{{M + \dfrac{m}{2}}}{{M + m}}} \right)(From equation 1)
∵(M+m2)<(M+m)\because \left( {M + \dfrac{m}{2}} \right) < \left( {M + m} \right)
∴v3>v\therefore {v_3} > v
Thus, they will impact greater velocity to the carriage by jumping one after the other.
Hence, option B is correct .

Note: In the solution, we used the concept of the conservation of momentum. The law of conservation of momentum states that for a system the momentum is always conserved when no external force is applied. It means for a system of two bodies if one body gets momentum then the other body will get an equal momentum but in the opposite direction.