Question

A 0.15 kg marble sliding to the right at 2.5 m/s on a frictionless surface makes a head-on collision with a 0.25 kg marble moving

to the left at 3.0 m/s. After the collision, the first marble moves to the left at 2.8 m/s. How fast does the second marble move to
the right?

Answer 1

The final velocity of the second marble is -1.5 m/s.

Step-by-step explanation:

In this problem, we can apply the principle of conservation of momentum to solve for the final velocity of the second marble. Before the collision, the total momentum of the system is zero since the marbles are moving in opposite directions. After the collision, the first marble moves to the left at 2.8 m/s. Assuming the second marble moves to the right, we can use the equation:

m1v1 + m2v2 = m1v'1 + m2v'2

Where m1 and m2 are the masses of the marbles, v1 and v2 are their initial velocities, and v'1 and v'2 are their final velocities. Plugging in the values given:

(0.15 kg)(2.5 m/s) + (0.25 kg)(-3.0 m/s) = (0.15 kg)(-2.8 m/s) + (0.25 kg)(v'2)

Simplifying the equation:

0.375 - 0.75 = -0.42 + 0.25v'2

-0.375 = 0.25v'2

v'2 = -1.5 m/s

Therefore, the second marble moves to the left at a velocity of 1.5 m/s.

Answer 2

Answer: 0.18 m/s

Step-by-step explanation:

This is a conservation of momentum problem. Total momentum before the collision must equal the total momentum after the collision.

Momentum = p = mv

let x = velocity of the second marble after the collision

(0.15 kg)(2.5 m/s) + (0.25 kg)(-3.0 m/s) = (0.15 kg)(-2.8 m/s) + (0.25 kg)(x)

-0.375 kg·m/s = -0.42 kg·m/s + (0.25 kg)x

(0.25 kg)x = (0.42 - 0.375) kg·m/s = 0.045 kg·m/s

x = (0.045 kg·m/s) / (0.25 kg) = 0.18 m/s