Question

Someone drops a brick on a 2.6 kg cart moving at 28.2 cm/s. After the collision, the dropped brick and cart are moving together with a velocity of 15.7 cm/s. Determine the mass of the dropped brick.

Answer 1

Answer:

2.087 kg

Explanation:

From the law of conservation of momentum,

Total momentum before collision = total momentum after collision.

MU + mu = V(M+m)............................ Equation 1

Where M = mass of  the cart, m = mass of the brick, U = initial velocity of the cart, u = initial velocity of the brick, V = velocity of both cart and brick after collision.

Note: The initial velocity of the brick is zero Thus mu = 0

making m the subject of the equation,

m =(MU/V)-M ........................................ Equation 2

Given: M = 2.6 kg, U = 28.2 cm/s = 0.283 m/s, V = 15.7 cm/s = 0.157 m/s.

Substituting into equation 2,

m = (2.6×0.283/0.157)-2.6

m = 2.087 kg

Thus the mass of the brick = 2.087 kg