Ask Question
17 December, 21:53

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.

+2
Answers (1)
  1. 17 December, 22:20
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers