Question

A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 26 cm/s to the left. What is the velocity of the 4.5-kg cart? a. 10 cm/s b. 4.56 cm/s c. 8.67 cm/s d. 7.3 cm/s

Answer 1

Answer:

c. 8.67 cm/s

Explanation:

From the law of conservation of momentum,

Total momentum before the thread was burned = Total momentum after was burned

mu + m'u' = mv + m'v'...................... Equation

Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.

Note: Both cart where momentarily at rest, as  such u = u' = 0. i.e the total momentum before the thread was burn = 0

And assuming the left is positive,

We can rewrite equation 1 as

mv + m'v' = 0............................................ Equation 2

Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s

Substitute into equation 2,

4.5v + 1.5(26) = 0

4.5v + 39 = 0

4.5v = -39

v = -39/4.5

v = -8.67 cm/s.

Note: v is negative because it moves to right.

Hence the velocity of the 4.5 kg cart = 8.67 cm/s.

The right option is c. 8.67 cm/s