4. a 1200 kg car accelerates from rest to 10.0 m/s in a time of 4.50 seconds. calculate the force that the car’s tires exerted on the road. (5 pts)
5. define the following terms:
a. elastic collision (2 pts)
b. inelastic collision (2 pts)
c. a 2.50 kg ball moving at 7.50 m/s is caught by a 68 kg boy while the boy is standing on ice.
i) how fast will the boy/ball combination be moving after the ball is caught by the boy? (5 pts)
ii) is the collision elastic or inelastic? (1 pt)

Question

4. a 1200 kg car accelerates from rest to 10.0 m/s in a time of 4.50 seconds. calculate the force that the car’s tires exerted on the road. (5 pts)
5. define the following terms:
   a. elastic collision (2 pts)
   b. inelastic collision (2 pts)
   c. a 2.50 kg ball moving at 7.50 m/s is caught by a 68 kg boy while the boy is standing on ice.
      i) how fast will the boy/ball combination be moving after the ball is caught by the boy? (5 pts)
      ii) is the collision elastic or inelastic? (1 pt)

Accepted Answer

### Question 4
# Explanation:
## Step1: Find acceleration
Acceleration $ a = \frac{v - u}{t} $, where $ u = 0 \, \text{m/s} $, $ v = 10.0 \, \text{m/s} $, $ t = 4.50 \, \text{s} $
$ a = \frac{10.0 - 0}{4.50} = \frac{10.0}{4.50} \approx 2.222 \, \text{m/s}^2 $

## Step2: Calculate force
Using Newton's second law $ F = ma $, where $ m = 1200 \, \text{kg} $, $ a \approx 2.222 \, \text{m/s}^2 $
$ F = 1200 \times 2.222 \approx 2666.67 \, \text{N} $

# Answer:
The force exerted by the car's tires on the road is approximately $ \boldsymbol{2667 \, \text{N}} $ (or $ 2.67 \times 10^3 \, \text{N} $)

### Question 5a
# Brief Explanations:
An elastic collision is a collision where both momentum and kinetic energy are conserved. The total kinetic energy before the collision equals the total kinetic energy after the collision, and the total momentum is also conserved. Objects involved in an elastic collision bounce off each other without any permanent deformation or heat generation (ideal case, e.g., collisions between billiard balls in theory).

# Answer:
An elastic collision is a collision in which both momentum and kinetic energy are conserved (objects bounce off each other, and total kinetic energy before = after collision).

### Question 5b
# Brief Explanations:
In an inelastic collision, momentum is conserved, but kinetic energy is not. Some kinetic energy is lost (converted to heat, sound, or used to deform objects). Often, the colliding objects stick together (perfectly inelastic collision) or move with a common velocity after the collision.

# Answer:
An inelastic collision is a collision where momentum is conserved, but kinetic energy is not (kinetic energy is lost, e.g., objects stick together after collision).

### Question 5c i
# Explanation:
## Step1: Apply conservation of momentum
Initial momentum $ p_{\text{initial}} = m_{\text{ball}} v_{\text{ball}} + m_{\text{boy}} v_{\text{boy}} $ (boy is initially at rest, so $ v_{\text{boy}} = 0 $)
Final momentum $ p_{\text{final}} = (m_{\text{ball}} + m_{\text{boy}}) v_{\text{final}} $
By conservation of momentum: $ m_{\text{ball}} v_{\text{ball}} = (m_{\text{ball}} + m_{\text{boy}}) v_{\text{final}} $

## Step2: Solve for $ v_{\text{final}} $
Given $ m_{\text{ball}} = 2.50 \, \text{kg} $, $ v_{\text{ball}} = 7.50 \, \text{m/s} $, $ m_{\text{boy}} = 68 \, \text{kg} $
$ v_{\text{final}} = \frac{m_{\text{ball}} v_{\text{ball}}}{m_{\text{ball}} + m_{\text{boy}}} = \frac{2.50 \times 7.50}{2.50 + 68} $
$ v_{\text{final}} = \frac{18.75}{70.5} \approx 0.266 \, \text{m/s} $

# Answer:
The boy/ball combination will be moving at approximately $ \boldsymbol{0.266 \, \text{m/s}} $ (or $ 0.267 \, \text{m/s} $) after the collision.

### Question 5c ii
# Brief Explanations:
To check if the collision is elastic, compare initial and final kinetic energy.
Initial KE: $ KE_{\text{initial}} = \frac{1}{2} m_{\text{ball}} v_{\text{ball}}^2 = \frac{1}{2} (2.50)(7.50)^2 \approx 70.31 \, \text{J} $
Final KE: $ KE_{\text{final}} = \frac{1}{2} (m_{\text{ball}} + m_{\text{boy}}) v_{\text{final}}^2 = \frac{1}{2} (70.5)(0.266)^2 \approx 2.47 \, \text{J} $
Since $ KE_{\text{initial}}
eq KE_{\text{final}} $, kinetic energy is not conserved. Thus, the collision is inelastic (the ball and boy stick together, which is a sign of inelastic collision).

# Answer:
The collision is inelastic (kinetic energy is not conserved, and the ball and boy move together after the collision).

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QUESTION IMAGE

Question

Explanation:

Question 4

Step1: Find acceleration

Step2: Calculate force

Answer:

Question 5a