A wheeled cart with a mass of 0.50 kg is rolling along a horizontal track at a c

Question

A wheeled cart with a mass of 0.50 kg is rolling along a horizontal track at a constant velocity of 2.0 m/s when it experiences an elastic collision with a second identical cart that is initially at rest, and attached to a spring with a spring constant of 4.0 N/m. After the collision, the second cart moves to the right. What is the first cart doing after the collision? Briefly justify your answer.

Explanation / Answer

f=20Hz T=1/f =1/20 =0.05s For the length we can use the formula: T= 2pi [v(L/32)] =>L=9.81(T/2pi)^2 =9.81(0.05/2*3.14)^2 =0.00062m =0.62cm Second question: a) Since the cart is moving in a constant velocity => it is in URM or uniform rectilinear motion b)Although I don't have the figure, I am going to assume that cart 1 strikes cart 2 after it finishes the incline and comes to a halt after collision. Moreover we are going to assume elastic collision. This is a linear momentum case, under which linear momentum is conserved P1+P2=P3+P4 where P1 and P2 are the momenta of cart 1 and cart 2 before collision respectively and, P3 and P4 are the momenta of cart 1 and cart 2 after collision respectively P1=M1V1 P2=M2V2=0 since cart 2 was initially at rest P3=M3V3=0 since cart 1 stops P4=M4V4 M1=M2=M3=M4 =>M1V1=M4V4 since they are identical carts => they have same mass =>M1=M4 =>V1=V4=2m/s Energy that the cart has before compression of the spring is kinetic =1/2mv^2 =0.5*0.5*2^2 =1J This will all be transformed to elastic energy of the spring =1/2kx^2 =>1/2kx^2=1 =>x=v2/k =>x=v2/4 =>x=v2/2m=0.707m c)There will be a second collision since the spring will perform an opposite and equal reaction on the cart Cart 1 will be performing uniformly accelerated motion in the opposite direction. Cart 2 will be at rest. Mind u that due to the lack of the figure 12.24, I assumed that the collision occurred on a horizontal surface If it is to occur on the incline the whole problem is changed since by now we would have the two carts combine after collision so the mass will double in the equation M1V1=M4V4 and it would become M1V1=M3V3+M4V4 where V3=V4 b)part b becomes =>M1V1=(M3+M4)V =>V=M1V1/(M3+M4) =>V=MV1/2M =>V=V1/2=1m/s Thus Energy that the cart has before compression of the spring is kinetic =1/2mv^2 =0.5*0.5*1^2 =0.25J Thus This will all be transformed to elastic energy of the spring =1/2kx^2 =>1/2kx^2=0.25 =>x=v1/2k =>x=v1/8 =>x=v2/4m=0.354m c) and part c becomes Both cart 1 and cart 2 would be performing uniformly accelerated motion followed by uniformly decelerated motion.

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