Question

A 40 g marble moving at 2.1 m/s strikes a 28 g marble at rest. Assume the collision is perfectly elastic and the marbles collide head on.

What is the speed of the first marble immediately after the collision?

Answer 1

To find the final speed of the first marble in an elastic collision, we must use the principles of conservation of momentum and conservation of kinetic energy to set up and solve a system of equations. By substituting given values into the formulas, we can solve for the final speed of the first marble after the collision.

Step-by-step explanation:

The problem involves a perfectly elastic collision between two marbles of different masses. For an elastic collision, both momentum and kinetic energy are conserved. Using these principles, we can set up two equations: one for the conservation of momentum and one for the conservation of kinetic energy.

Conservation of momentum before and after the collision for the two marbles can be represented by:

m1⋅v1_initial + m2⋅v2_initial = m1⋅v1_final + m2⋅v2_final

Since the second marble is initially at rest v2_initial is zero:

40g⋅2.1m/s + 0 = 40g⋅v1_final + 28g⋅v2_final

For the conservation of kinetic energy, the equation is:

0.5⋅m1⋅v1_initial^2 = 0.5⋅m1⋅v1_final^2 + 0.5⋅m2⋅v2_final^2

Solving these equations simultaneously will give us the final velocities of both marbles. However, the student only asked for the speed of the first marble, which we'll denote as v1_final.