Question

Crall and Whipple observe two carts collide elastically in a laboratory experiment. Cart 1 has a mass of 1.510 kg, and cart 2 has a mass of 1.190 kg. Crall chooses to work in the laboratory frame. Before the collision in his frame, the carts' speeds are (v_1i). What is the speed of cart 2 after the collision in the laboratory frame?

a) (v_2f = 2m_1m_1 + m_2 ⋅ v_1i)

b) (v_2f = m_1 - m_2m_1 + m_2 ⋅ v_1i)

c) (v_2f = 2m_2m_1 + m_2 ⋅ v_1i)

d) (v_2f = m_2 - m_1m_1 + m_2 ⋅ v_1i)

Answer 1

The correct formula for the final speed of cart 2 after an elastic collision, with one cart initially at rest, is v_2f = (2m_1)/(m_1 + m_2) x v_1i, which corresponds to option c).

Step-by-step explanation:

In an elastic collision, both momentum and kinetic energy are conserved. The final velocity of cart 2 in the laboratory frame can be determined using the equations derived from the conservation of momentum and kinetic energy. Given that the initial speed of cart 2 is not specified, we can assume it to be at rest if not stated otherwise. The formula for the final velocity of cart 2 after an elastic collision when one object is initially at rest is v_2f = (2m_1)/(m_1 + m_2) × v_1i. Substituting the given masses, we obtain v_2f = (2 × 1.510kg)/(1.510kg + 1.190kg) × v_1i.

Therefore, the correct formula for the final speed of cart 2 after the collision in the provided options is:

c) (v_2f = 2m_1)/(m_1 + m_2) × v_1i