Im not sure how to do this question without the masses given

Question

Pendulum A is released from a height of 2.0 m above its lowest point. It swings and, at the moment when the string is vertical, the pendulum collides with block B, which is initially at rest and has double the mass as ball A. The collision is head-on and elastic, and the floor is frictionless. Determine the velocity of B after the collision.

Answer 1

The mass is not given because the mass is not required to solve this problem. Instead, they give you a ratio of masses (ball B has double the mass as ball A... in other words, m_b = 2*m_a).

To solve this problem, you first need to get the velocity of ball A the moment before it hits ball B.

You can use the conservation of energy to solve this first part (E_initial = E_final). The potential energy of ball A on the pendulum is equal to the kinetic energy at the bottom of the swing. where potential energy U = m*g*h, and kinetic energy K = (1/2)*m*v^2).

U = K

(1/2)*m_a * v_a_initial^2 = m_a *g*h

Notice the masses cancel. Solve for v_a_initial.

Now, to solve for the velocity of ball B after the collision (v_b_final), use the conservation of momentum (p_initial = p_final, where p = m*v).

The equation should include the momentum of both balls right before the collision on the left side and the momentum of both balls right after the collision on the right side. The velocity of ball B before the collision is zero, so ball B has no momentum.

(m_a * v_a_initial) + (m_b *0) = (m_a * v_a_final) + (m_b * v_b_final)

There are two unknowns in the equation: v_a_final and v_b_final. We need another equation in order to solve it. Since the collision is elastic, the kinetic energy is conserved (K_initial = K_final)

((1/2)*m_a * v_a_initial^2) + (m_b *0) = ((1/2)*m_a * v_a_final^2) + ((1/2)*m_b * v_b_final^2)

When we use the mass ratio given in the problem, the masses will cancel in both the momentum and kinetic energy equations. Algebraically rearrange both equations to get v_b_final by itself.