Question

1. cart 1 (m = 9.0 kg) accelerates from 1.0 m/s to 2.5 m/s after experiencing an elastic collision with cart 2 (m = 4.0 kg).

1a: what is cart 1’s change in momentum?
| known | m =
vi =
vf = |

unknown

δp

| tool box | momentum
p = (m)(v)

inelastic collision
m1vi + m2vi = vf(m1 + m2)

elastic collision
m1vi + m2vi = m1vf + m2vf

change in momentum
δp = (mv)f - (mv)i
δp = m(vf - vi) |

substitute and solve

1b: what is cart 2’s change in momentum? explain how you know. (clue: go back and read the stamp)

Explanation:

1A Solution:

Step1: Identify known values

Mass of cart 1, \( m = 9.0 \, \text{kg} \), initial velocity \( v_i = 1.0 \, \text{m/s} \), final velocity \( v_f = 2.5 \, \text{m/s} \).

Step2: Use change in momentum formula

The formula for change in momentum is \( \Delta p = m(v_f - v_i) \).

Step3: Substitute values

Substitute \( m = 9.0 \, \text{kg} \), \( v_f = 2.5 \, \text{m/s} \), \( v_i = 1.0 \, \text{m/s} \) into the formula:
\( \Delta p = 9.0 \times (2.5 - 1.0) \)

Step4: Calculate

First, find \( 2.5 - 1.0 = 1.5 \, \text{m/s} \). Then, \( \Delta p = 9.0 \times 1.5 = 13.5 \, \text{kg·m/s} \).

In an elastic collision, the total momentum of the system is conserved. This means the change in momentum of cart 1 (\( \Delta p_1 \)) and the change in momentum of cart 2 (\( \Delta p_2 \)) must satisfy \( \Delta p_1 + \Delta p_2 = 0 \) (since the total momentum before and after the collision is the same, so the net change in momentum of the system is zero). From part 1A, \( \Delta p_1 = 13.5 \, \text{kg·m/s} \), so \( \Delta p_2 = -\Delta p_1 = -13.5 \, \text{kg·m/s} \). The negative sign indicates cart 2’s momentum change is opposite in direction to cart 1’s, but the magnitude is equal (due to conservation of momentum in elastic collisions).

Answer:

The change in momentum of cart 1 is \( 13.5 \, \text{kg·m/s} \).

1B Solution: