Question

1. cart 1 (m = 14 kg) is moving with a velocity of +1.5 m/s when it collides inelastically with cart 2 (m = 13 kg). cart 2 was moving with a velocity of +0.25 m/s before the collision.

2a. calculate the velocity of both carts after the collision.

| known | cart 1
m₁ =
vᵢ =

cart 2
m₂ =
vᵢ = |

unknown

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

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

elastic collision
m₁vᵢ + m₂vᵢ = m₁v_f + m₂v_f

change in momentum
δp = (mv)_f - (mv)_i
δp = m(v_f - v_i) |

substitute and solve

Explanation:

Step1: Identify known values

For Cart 1: \( m_1 = 14 \, \text{kg} \), \( v_{i1} = 1.5 \, \text{m/s} \)
For Cart 2: \( m_2 = 13 \, \text{kg} \), \( v_{i2} = 0.25 \, \text{m/s} \)
Use the inelastic collision formula: \( m_1v_{i1} + m_2v_{i2} = v_f(m_1 + m_2) \)

Step2: Substitute values into the formula

Left side (total initial momentum):
\( (14)(1.5) + (13)(0.25) = 21 + 3.25 = 24.25 \, \text{kg·m/s} \)

Right side (total mass times final velocity):
\( v_f(14 + 13) = v_f(27) \)

Step3: Solve for \( v_f \)

\( 27v_f = 24.25 \)
\( v_f = \frac{24.25}{27} \approx 0.898 \, \text{m/s} \) (rounded to three decimal places)

Answer:

The velocity of both carts after the collision is approximately \( \boldsymbol{0.90 \, \text{m/s}} \) (or \( \approx 0.898 \, \text{m/s} \))