Question

A 250 g blob of clay moves with a speed of 15 m/s towards a 400 g cart that is initially at rest. a). What is the momentum of the system before the blob of clay strikes the cart? b). What must be the momentum of the system after they come together? ( Please note, a response of "the same " is insufficient and will result in minimal credit). c. If the blob of clay sticks to the cart, with what speed will the clay and cart move after they come together?

Answer 1

Hi there!

a)

We can use the following equation to solve for the momentum. Remember that momentum is ALWAYS CONSERVED.

p = m · v

Plug in the given values:

p = (.250) · 15 = 3.75 kgm/s

b)

The momentum of the system AFTER the collision will be the same because of a CONSERVATION OF MOMENTUM.

$p_i = p_f$

c)

We know that by the conservation of momentum:

m₁v₁ + m₂v₂ = m₁v₁ '+ m₂v₂'

The second object (cart) is originally at rest, and the two objects move together after the collision, so:

m₁v₁ = (m₁ + m₂)vf

Solve by plugging in values:

(.250)(15)/(.250 + .400) = vf = 5.769 m/s