Question

A 2.4-kg cart is rolling along a frictionless, horizontal track towards a 1.7-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is 4.1 m/s, and the second cart's velocity is -2.8 m/s. (a) What is the total momentum of the system of the two carts at this instant

Answer 1

Step-by-step explanation:

Mass of first cart M1=2.4kg

Velocity of first cart U1=4.1m/s

Mass of second cart M2=1.7kg

Second cart is initially at rest U2=0

After an instant, the velocity of the second cart is U2=-2.8m/s

Now after collision the two cart move together with the same velocity I.e inelastic collision

Using conservation of momentum

Momentum before collision, = momentum after collision

M1U1 + M2U2 = (M1+M2)V

2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V

9.84 - 4.76 = 4.1V

5.08=4.1V

V=5.08/4.1

V=1.24m/s

The momentum of the two cart at that instant is

M1U1+M2U2

2.4×4.1 + 1.7× -2.8

9.84 - 4.76

5.08kgm/s

So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s