Umm..to experts of momentum lesson..please help?

Question

Railroad car A rolls at a certain speed and makes a perfectly elastic collision with car B of the same mass. After the collition, car A is observed to be at rest. How does the speed of the car B compare to the initial speed of car A.

Okay now even though your not expert in physics....please help

please i don't need formulas I need explanations.....

Answer 1

If the collision is perfectly elastic (which is unlikely) and the cars are of the same mass - then car B's velocity will be the same as Car A's was.

The reason the collision won't be perfectly elastic is that you would hear the cars collide thus giving off energy as sound and also heat would be generated during the collision due to friction.

Answer 2

The final speed of car B equals the initial speed of car A.

> please i don't need formulas I need explanations

Sorry, Mu: In most of physics, the formula IS the explanation.

Proof:

By conservation of momentum:
(total momentum before) = (total momentum after)
(m_a)(v_a_initial) + (m_b)(v_b_initial) = (m_a)(0) + (m_b)(v_b_final)

Since the problem states that m_a = m_b, we'll use m_a everywhere. Then:
(m_a)(v_a_initial) + (m_a)(v_b_initial) = (m_a)(v_b_final)

or:
v_a_initial + v_b_initial = v_b_final

Next, we use the fact that the collision is elastic:
(total KE before) = (total KE after)
(m_a)(v_a_initial)²/2 + (m_b)(v_b_initial)²/2 = (m_a)(0)²/2 + (m_b)(v_b_final)²/2

Again using m_a = m_b:
(m_a)(v_a_initial)²/2 + (m_a)(v_b_initial)²/2 = (m_a)(v_b_final)²/2

or:
(v_a_initial)² + (v_b_initial)² = (v_b_final)²

We showed previously that v_b_final = v_a_initial + v_b_initial. So, substituting:
(v_a_initial)² + (v_b_initial)² = (v_a_initial+v_b_initial)²
= (v_a_initial)² + 2(v_a_initial)(v_b_initial) + (v_b_initial)²

From which:
2(v_a_initial)(v_b_initial) = 0

This means that either v_a_initial or v_b_initial (or both) are zero. But we are assuming that car A was initially rolling, so v_a_initial is not zero; therefore:

v_b_initial = 0

Combining this with (v_b_final = v_a_initial + v_b_initial), we get:

v_b_final = v_a_initial

Answer 3

In elastic collisions, both kinetic energy and momentum are conserved. So

Initial momentum = pi = mAvA

Final momentum = pf = mB vB

pi = pf ---> mA vA = mB vB ---> vB = vA* (mA/mB)

Answer 4

car a moved moved 50.1693 % compared 2 car b 3=x%90.490+y=?