A cart of mass 0.400 kg moves with a speed of 1.2 m/s toward a secondary cart of mass 0.300 kg that is initially at rest. When the carts collide they stick together. What speed do the carts move with after the collision

0.686 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V (m+m') ... Equation 1

Where m = mass of the cart, m' = mass of the secondary cart, u = initial speed of the cart, u' = initial speed of the secondary cart, V = common speed after collision.

make V the subject of the equation

V = (mu+m'u') / (m+m') ... Equation 2

Given: m = 0.4 kg, u = 1.2 m/s, m' = 0.3 kg, u' = 0 m/s (at rest)

Substitute into equation 2

V = (0.4*1.2+0.3*0.3*0) / (0.4+0.3)

V = 0.48/0.7

V = 0.686 m/s

0.686 m/s

Explanation:

The principle of conservation says the sum of momenta before collision is equal to the sum of momenta after collision in an isolated system. Here, we assume the system is not acted upon by any external force.

The initial momentum = momentum of 1 st cart + momentum of 2nd cart

= 0.400 * 1.2 + 0.300 * 0 = 0.48 kg m/s

The final momentum = (0.400 + 0.300) * v (since they stick together)

0.700 v = 0.48

v = 0.686 m/s