Impulse & Momentum Worksheets with illustrated examples of physics problems.

Let's solve each part of the Impulse & Momentum Worksheet step by step. We'll use the following key physics concepts:

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🔹 Key Formulas:

- Momentum (p) = mass × velocity → $ p = m \cdot v $
- Impulse (J) = change in momentum → $ J = \Delta p = m(v_f - v_i) $
- Impulse is also equal to force × time, but here we're focusing on momentum changes.

We’ll go through each scenario (A, B, C, D) and calculate what’s asked.

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## ✔ Problem A: Bike Rider

Given:
- Mass of bike ($ M_b $) = 60 kg
- Mass of rider ($ M_r $) = 20 kg
- Velocity ($ v $) = 5 m/s (both moving together)

Step 1: Momentum of the bike

$$
p_{\text{bike}} = M_b \cdot v = 60 \, \text{kg} \times 5 \, \text{m/s} = 300 \, \text{kg·m/s}
$$

Step 2: Momentum of the rider

$$
p_{\text{rider}} = M_r \cdot v = 20 \, \text{kg} \times 5 \, \text{m/s} = 100 \, \text{kg·m/s}
$$

Step 3: Impulse?

There is no change in velocity mentioned — the system is moving at constant velocity. So:
- Impulse = 0, because impulse is change in momentum.
- No external force or collision described → no change in motion.

✔ Answer for A:
- Impulse: 0
- Momentum of the bike: 300 kg·m/s
- Momentum of the rider: 100 kg·m/s

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## ✔ Problem B: Calvin catches a ball

Given:
- Ball initial velocity ($ v_0 $) = 30 m/s
- Mass of ball ($ m $) = 0.150 kg
- Calvin catches the ball → assume he stops it (final velocity = 0 m/s)
- Assume Calvin was initially at rest.

We are to find impulse on the ball (or on Calvin — same magnitude, opposite direction).

Step 1: Change in momentum of the ball

$$
\Delta p = m(v_f - v_i) = 0.150 \, \text{kg} \times (0 - 30) \, \text{m/s} = -4.5 \, \text{kg·m/s}
$$

So the impulse on the ball is -4.5 N·s (negative means opposite to initial direction).

But impulse magnitude is often asked as absolute value unless direction matters.

✔ Answer for B:
- Impulse on the ball: $ \boxed{-4.5} \, \text{N·s} $ (or 4.5 N·s opposite to motion)
- This is also the impulse experienced by Calvin (equal and opposite).

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## ✔ Problem C: Baseball hit back

Before:
- Ball mass $ m = 0.150 \, \text{kg} $
- Initial velocity $ v_i = -20 \, \text{m/s} $ (toward pitcher, so negative)
- Final velocity $ v_f = +35 \, \text{m/s} $ (away from pitcher, positive)

Step 1: Change in momentum (impulse)

$$
\Delta p = m(v_f - v_i) = 0.150 \times [35 - (-20)] = 0.150 \times 55 = 8.25 \, \text{kg·m/s}
$$

So impulse on the ball is +8.25 N·s (in the direction away from pitcher).

✔ Answer for C:
- Impulse on the ball: $ \boxed{8.25} \, \text{N·s} $

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## ✔ Problem D: Snowball hits Susie and sticks

Given:
- Snowball mass $ m = 0.1 \, \text{kg} $
- Initial velocity $ v_i = 10 \, \text{m/s} $
- It sticks to Susie → final velocity = 0 m/s (assume Susie doesn't move much, so snowball stops)

Step 1: Change in momentum of snowball

$$
\Delta p = m(v_f - v_i) = 0.1 \times (0 - 10) = -1.0 \, \text{kg·m/s}
$$

So impulse on the snowball is -1.0 N·s

But impulse on Susie is +1.0 N·s (equal and opposite)

✔ Answer for D:
- Impulse on the snowball: $ \boxed{-1.0} \, \text{N·s} $
- Or if asking for impulse *on Susie*, it's $ +1.0 \, \text{N·s} $

But since the question says “the snowball hits Susie and sticks”, likely asking for impulse delivered to Susie (which is equal to the change in momentum of the snowball, with opposite sign).

So:
- Impulse on Susie: $ \boxed{1.0} \, \text{N·s} $

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✔ Final Answers Summary:

#### A
- Impulse: 0
- Momentum of the bike: 300 kg·m/s
- Momentum of the rider: 100 kg·m/s

#### B
- Impulse: -4.5 N·s (or magnitude 4.5 N·s, direction opposite to motion)

#### C
- Impulse: 8.25 N·s (positive, toward the batter)

#### D
- Impulse (on Susie): 1.0 N·s

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