1. Two ice skaters skate on a slippery (frictionless) ice rink. Daniel (mass 60 kg ) is skating northwards at $$ 1 \mathrm{~m} . \mathrm{s}-1 $$ and Lauren (mass 50 kg ) is skating towards him at $$ 0,6 \mathrm{~m} . \mathrm{s}-1 $$. They collide inelastically with each other. Lauren's velocity after the collision is $$ 0,8 \mathrm{~m} . \mathrm{s}- $$ l, north. a) Calculate Daniel's velocity after the collision. b) Calculate the change in kinetic energy during the collision.

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Let's analyze the collision between Daniel and Lauren step by step.

### **Given:**
- **Daniel:**
  - Mass ($$ m_1 $$) = 60 kg
  - Initial velocity ($$ v_{1i} $$) = $$ +1 \, \text{m/s} $$ (northwards)
- **Lauren:**
  - Mass ($$ m_2 $$) = 50 kg
  - Initial velocity ($$ v_{2i} $$) = $$ -0.6 \, \text{m/s} $$ (southwards, since she's moving towards Daniel)
- **After Collision:**
  - Lauren's velocity ($$ v_{2f} $$) = $$ +0.8 \, \text{m/s} $$ (northwards)

### **a) Calculating Daniel's Velocity After the Collision ($$ v_{1f} $$):**

Since the ice is frictionless, **momentum is conserved** during the collision. The conservation of momentum equation is:

$$
m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}
$$

Plugging in the known values:

$$
60 \times 1 + 50 \times (-0.6) = 60 \times v_{1f} + 50 \times 0.8
$$

$$
60 - 30 = 60 v_{1f} + 40
$$

$$
30 = 60 v_{1f} + 40
$$

$$
60 v_{1f} = 30 - 40 = -10
$$

$$
v_{1f} = \frac{-10}{60} = -\frac{1}{6} \, \text{m/s}
$$

**Interpretation:** Daniel's velocity after the collision is $$ -\frac{1}{6} \, \text{m/s} $$, which means he moves southwards at $$ \frac{1}{6} \, \text{m/s} $$.

### **b) Calculating the Change in Kinetic Energy ($$ \Delta K $$):**

**Initial Kinetic Energy ($$ K_i $$):**

$$
K_{1i} = \frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} \times 60 \times 1^2 = 30 \, \text{J}
$$
$$
K_{2i} = \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} \times 50 \times (-0.6)^2 = 0.5 \times 50 \times 0.36 = 9 \, \text{J}
$$
$$
K_i = K_{1i} + K_{2i} = 30 + 9 = 39 \, \text{J}
$$

**Final Kinetic Energy ($$ K_f $$):**

$$
K_{1f} = \frac{1}{2} m_1 v_{1f}^2 = \frac{1}{2} \times 60 \times \left(-\frac{1}{6}\right)^2 = 30 \times \frac{1}{36} = \frac{5}{6} \, \text{J} \approx 0.833 \, \text{J}
$$
$$
K_{2f} = \frac{1}{2} m_2 v_{2f}^2 = \frac{1}{2} \times 50 \times 0.8^2 = 25 \times 0.64 = 16 \, \text{J}
$$
$$
K_f = K_{1f} + K_{2f} = \frac{5}{6} + 16 = \frac{101}{6} \, \text{J} \approx 16.833 \, \text{J}
$$

**Change in Kinetic Energy:**

$$
\Delta K = K_f - K_i = \frac{101}{6} - 39 = \frac{101}{6} - \frac{234}{6} = -\frac{133}{6} \, \text{J} \approx -22.17 \, \text{J}
$$

**Interpretation:** The kinetic energy decreases by $$ \frac{133}{6} \, \text{J} $$ (approximately $$ -22.17 \, \text{J} $$) during the collision.

### **Final Answers:**

**a)** After the collision, Daniel’s velocity is –1⁄6 m s⁻¹ (i.e., 1/6 m/s south).

**b)** The kinetic energy decreases by 133⁄6 J during the collision.
After the collision, Daniel moves south at 1/6 m/s. The kinetic energy decreases by 133⁄6 J during the collision.

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