Question

1. while roller - skating, granny collides with her tiny grandson, ambrose, who is at rest. ignoring any friction effects, ambroses speed after the collision will be greatest when...

a. granny catches him, and they both move together.
b. he and granny make a bouncing collision, each going in opposite directions.

short answer (1 question, 9 points)
momentum:
p = mv

1. impulse:

j = fδt

1. impulse - momentum theorem:

j = δp
fδt = mδv

1. conservation of momentum:

p_b = p_a b = before, a = after

1. elastic collision:

m_1v_1b + m_2v_2b = m_1v_1a + m_2v_2a

1. inelastic collision:

m_1v_1b + m_2v_2b=(m_1 + m_2)v_a

1. a 1,500 kg car is stalled out in the middle of a railroad crossing. an oncoming 10,000 kg locomotive traveling at 10 m/s is unable to stop and collides with the car. the car and train stick together and skid down the tracks.

a. draw a diagram of the collision. include masses and velocities in your diagram. (2 points)

Explanation:

To solve for the velocity of the combined car and locomotive after the collision, we use the principle of conservation of momentum for an inelastic collision (since they stick together).

Step 1: Identify the given values

• Mass of the car, \( m_1 = 1500 \, \text{kg} \) (initially at rest, so \( v_{1B} = 0 \, \text{m/s} \))
• Mass of the locomotive, \( m_2 = 10000 \, \text{kg} \)
• Velocity of the locomotive before collision, \( v_{2B} = 10 \, \text{m/s} \)
• Let \( v_A \) be the velocity of the combined mass after the collision.

Step 2: Apply the inelastic collision formula

The formula for conservation of momentum in an inelastic collision is:

$$ m_1v_{1B} + m_2v_{2B} = (m_1 + m_2)v_A $$

Step 3: Substitute the known values into the formula

Since \( v_{1B} = 0 \, \text{m/s} \), the equation simplifies to:

$$ 0 + (10000 \, \text{kg})(10 \, \text{m/s}) = (1500 \, \text{kg} + 10000 \, \text{kg})v_A $$

Step 4: Calculate the left - hand side and the total mass

• Left - hand side: \( (10000 \, \text{kg})(10 \, \text{m/s})=100000 \, \text{kg}\cdot\text{m/s} \)
• Total mass: \( m_1 + m_2=1500 + 10000 = 11500 \, \text{kg} \)

Step 5: Solve for \( v_A \)

$$ v_A=\frac{100000 \, \text{kg}\cdot\text{m/s}}{11500 \, \text{kg}}\approx 8.7 \, \text{m/s} $$

The velocity of the combined car and locomotive after the collision is approximately \(\boldsymbol{8.7 \, \text{m/s}}\) (in the direction of the locomotive's initial motion).

Answer:

To solve for the velocity of the combined car and locomotive after the collision, we use the principle of conservation of momentum for an inelastic collision (since they stick together).

Step 1: Identify the given values

• Mass of the car, \( m_1 = 1500 \, \text{kg} \) (initially at rest, so \( v_{1B} = 0 \, \text{m/s} \))
• Mass of the locomotive, \( m_2 = 10000 \, \text{kg} \)
• Velocity of the locomotive before collision, \( v_{2B} = 10 \, \text{m/s} \)
• Let \( v_A \) be the velocity of the combined mass after the collision.

Step 2: Apply the inelastic collision formula

The formula for conservation of momentum in an inelastic collision is:

$$ m_1v_{1B} + m_2v_{2B} = (m_1 + m_2)v_A $$

Step 3: Substitute the known values into the formula

Since \( v_{1B} = 0 \, \text{m/s} \), the equation simplifies to:

$$ 0 + (10000 \, \text{kg})(10 \, \text{m/s}) = (1500 \, \text{kg} + 10000 \, \text{kg})v_A $$

Step 4: Calculate the left - hand side and the total mass

• Left - hand side: \( (10000 \, \text{kg})(10 \, \text{m/s})=100000 \, \text{kg}\cdot\text{m/s} \)
• Total mass: \( m_1 + m_2=1500 + 10000 = 11500 \, \text{kg} \)

Step 5: Solve for \( v_A \)

$$ v_A=\frac{100000 \, \text{kg}\cdot\text{m/s}}{11500 \, \text{kg}}\approx 8.7 \, \text{m/s} $$

The velocity of the combined car and locomotive after the collision is approximately \(\boldsymbol{8.7 \, \text{m/s}}\) (in the direction of the locomotive's initial motion).