Impulse/Velocity 3

Question

Two objects travel towards each other and undergo a perfectly elastic interaction. Before they collide, object A has a mass of 495g and a velocity of -2m/s, while object B has a mass of 664g and a velocity of -17.8m/s. After they collide, object A travels -20.104m/s. The interaction takes 0.52s from start to finish.

What is the final velocity of B?    unit   

What is the impulse by A on B?    unit   

What is the force by A on B?    unit   

What is the impulse by B on A?    unit   

What is the force by B on A?    unit   

Answer 1

To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.

Let's start by finding the initial momentum and kinetic energy of the system:

Initial momentum: p = m_A * v_A + m_B * v_B = (0.495 kg) * (-2 m/s) + (0.664 kg) * (-17.8 m/s) = -12.95 kg m/s

Initial kinetic energy: K = (1/2) * m_A * v_A^2 + (1/2) * m_B * v_B^2 = (1/2) * (0.495 kg) * (2 m/s)^2 + (1/2) * (0.664 kg) * (17.8 m/s)^2 = 296.97 J

Since the collision is perfectly elastic, the total kinetic energy of the system is conserved:

K_final = (1/2) * m_A * v_A'^2 + (1/2) * m_B * v_B'^2

where v_A' and v_B' are the final velocities of objects A and B, respectively.

Using the conservation of momentum, we can write:

p = m_A * v_A' + m_B * v_B'

Now we have two equations with two unknowns (v_A' and v_B'). We can solve for them by combining the two equations:

v_A' = (m_A - m_B) / (m_A + m_B) * v_A + 2 * m_B / (m_A + m_B) * v_B v_B' = 2 * m_A / (m_A + m_B) * v_A + (m_B - m_A) / (m_A + m_B) * v_B

Plugging in the values, we get:

v_A' = (0.495 kg - 0.664 kg) / (0.495 kg + 0.664 kg) * (-2 m/s) + 2 * (0.664 kg) / (0.495 kg + 0.664 kg) * (-17.8 m/s) = -28.416 m/s

v_B' = 2 * (0.495 kg) / (0.495 kg + 0.664 kg) * (-2 m/s) + (0.664 kg - 0.495 kg) / (0.495 kg + 0.664 kg) * (-17.8 m/s) = -11.284 m/s

Therefore, the final velocity of B is 11.284 m/s (positive because it's in the opposite direction of its initial velocity).

To find the impulse by A on B and the force by A on B, we can use the impulse-momentum theorem:

impulse = p_final - p_initial

where p_final and p_initial are the final and initial momenta of object B, respectively.

Using the values, we get:

impulse = m_B * v_B' - m_B * v_B = (0.664 kg) * (-11.284 m/s) - (0.664 kg) * (-17.8 m/s) = 6.374 kg m/s

The force by A on B can be found using the formula:

force = impulse / time

where time is the duration of the collision. Plugging in the values, we get:

force = 6.374 kg m/s / 0.52 s = 12.25 N

To find the impulse by B on A and the force by B on A, we can use the same formulas but with the values for object A:

impulse = m_A * v_A