Answer to Question #288564 in Physics for Bobward

Question #288564

Two men, of mass 100 kg each, stand on a cart of mass 300 kg. The cart can roll with negligible friction along a north-south track, and everything is initially at rest. One man runs towards the north and jumps off the cart at a speed of 5.0 m/s, relative to the cart. After he has jumped, the second man runs towards the south and jumps off the cart, again with a speed of 5.0 m/s relative to the cart.


Calculate the speed and direction of the cart after both men have jumped off. 


1
Expert's answer
2022-01-18T13:42:27-0500

Apply conservation of momentum to find the speed of the cart (mass M) and the second man (mass m) as the first one (mass m) jumped off:


0=mv+(m+M)uu=mvm+M, u=1005100+300=1.25 m/s (1.25 m/s South).0=mv+(m+M)u→u=-\dfrac{mv}{m+M},\\\space\\ u=-\dfrac{100·5}{100+300}=-1.25\text{ m/s (1.25 m/s South)}.


Again, apply conservation of momentum to find the speed of the cart (mass M) as the second man (mass m) jumped off:


(M+m)u=m(v)+Mvc, vc=(M+m)u+mvM, vc=(300+100)(1.25)+1005300=0 m/s.(M+m)u=m(-v)+Mv_c,\\\space\\ v_c=\dfrac{(M+m)u+mv}{M},\\\space\\ v_c=\dfrac{(300+100)(-1.25)+100·5}{300}=0\text{ m/s}.


The cart will stand still.


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