Question

6. A thread holds two carts together on a frictionless surface. A spring is compressed between the two carts. After the thread between the two carts is cut the cart on the left with a mass of 1.5 kg moves away with a velocity of 27 cm/s. What is the velocity of the cart on the right that has a mass of 4.5 kg?

Answer 1

Answer: -9 cm/s

Explanation: The velocity of the right cart can be determined using the conservation of momentum principle. According to this principle, the total momentum of the system before the collision is equal to the total momentum after the collision.

The total momentum before the collision is 0, since the carts are initially at rest.

After the thread is cut, the total momentum of the system is given by:

m1 * v1 + m2 * v2 = 0

where m1 is the mass of the left cart, m2 is the mass of the right cart, and v1 and v2 are their respective velocities after the collision.

Substituting the given values, we have:

1.5 * 27 + m2 * v2 = 0

Solving for v2, we get:

v2 = -27 * 1.5 / 4.5 = -9 cm/s

Therefore, the velocity of the right cart after the collision is -9 cm/s.