QUESTION 4 (Start on a now page.)
Dancers have to. leam many skills, including how to land correctly. A dancer of mass
50 kg leaps Thto the air and lands feet first on the ground. She lands on the ground with
a velocity of . As she lands, she bends her knees and comes to a complete stop
in 0,2 seconds:
Calculate the momentum with which the dancer reaches the ground.
Define the term impulse of a force,
Calculate the magnitude of the net force acting on the dancer as she lands.
er knees.
Will the force now be GREATER THAN, SMALLER THAN or EQUAL TO the dancer performs the same jump as before but lands without bending
force calculated in QUESTION 4.3 ?
Give a reason for the answer to QUESTION 4.4 .

Ask by Chambers Ruiz. in South Africa

Jan 22,2025

Question

QUESTION 4 (Start on a now page.)
Dancers have to. leam many skills, including how to land correctly. A dancer of mass
50 kg leaps Thto the air and lands feet first on the ground. She lands on the ground with
a velocity of . As she lands, she bends her knees and comes to a complete stop
in 0,2 seconds:
Calculate the momentum with which the dancer reaches the ground.
Define the term impulse of a force,
Calculate the magnitude of the net force acting on the dancer as she lands.
er knees.
Will the force now be GREATER THAN, SMALLER THAN or EQUAL TO the dancer performs the same jump as before but lands without bending
force calculated in QUESTION 4.3 ?
Give a reason for the answer to QUESTION 4.4 .

Ask by Chambers Ruiz. in South Africa

Jan 22,2025

UpStudy AI · Accepted Answer

Sure, let's tackle each part of QUESTION 4 step by step.

---

### **4.1 Calculate the momentum with which the dancer reaches the ground.**

**Momentum (p)** is calculated using the formula:

$$
p = m \times v
$$

Where:
- $$ m = 50 \, \text{kg} $$ (mass of the dancer)
- $$ v = 5 \, \text{m/s} $$ (velocity upon landing)

**Calculation:**

$$
p = 50 \, \text{kg} \times 5 \, \text{m/s} = 250 \, \text{kg} \cdot \text{m/s}
$$

**Answer:**  
The momentum with which the dancer reaches the ground is **250 kg·m/s** downward.

---

### **4.2 Define the term impulse of a force, and calculate the magnitude of the net force acting on the dancer as she lands on her knees.**

**Definition of Impulse:**

**Impulse (J)** is a measure of the change in momentum of an object when a force is applied over a specific time interval. It is given by the formula:

$$
J = F \times \Delta t = \Delta p
$$

Where:
- $$ F $$ = Net force applied
- $$ \Delta t $$ = Time interval over which the force is applied
- $$ \Delta p $$ = Change in momentum

**Calculation of Net Force:**

Given:
- Initial momentum ($$ p_i $$) = 250 kg·m/s (from 4.1)
- Final momentum ($$ p_f $$) = 0 kg·m/s (since the dancer comes to a complete stop)
- Time interval ($$ \Delta t $$) = 0.2 seconds

1. **Determine the Change in Momentum ($$ \Delta p $$):**

$$
\Delta p = p_f - p_i = 0 - 250 = -250 \, \text{kg} \cdot \text{m/s}
$$

*(The negative sign indicates a change in direction, but since we are interested in magnitude, we'll use the absolute value.)*

$$
|\Delta p| = 250 \, \text{kg} \cdot \text{m/s}
$$

2. **Calculate the Net Force ($$ F $$):**

$$
F = \frac{|\Delta p|}{\Delta t} = \frac{250 \, \text{kg} \cdot \text{m/s}}{0.2 \, \text{s}} = 1250 \, \text{N}
$$

**Answer:**

- **Impulse Definition:**  
  Impulse is the product of the net force acting on an object and the time duration over which it acts, resulting in a change in the object's momentum.

- **Magnitude of Net Force:**  
  The magnitude of the net force acting on the dancer as she lands is **1250 Newtons (N)** upward.

---

### **4.4 Will the force now be GREATER THAN, SMALLER THAN, or EQUAL TO the force calculated in QUESTION 4.2? Give a reason for your answer.**

**Answer:**  
**The force will be **SMALLER THAN** the force calculated in QUESTION 4.2.**

**Reason:**  
When the dancer bends her knees while landing, she increases the time interval ($$ \Delta t $$) over which she comes to a complete stop. According to the impulse formula:

$$
F = \frac{\Delta p}{\Delta t}
$$

Increasing $$ \Delta t $$ results in a smaller force $$ F $$ for the same change in momentum ($$ \Delta p $$). Bending the knees effectively spreads out the deceleration process, reducing the peak force experienced by the body during landing.

---

**Summary:**
- **4.1:** Momentum upon landing is **250 kg·m/s** downward.
- **4.2:** Impulse is the change in momentum due to force over time, and the net force during landing is **1250 N** upward.
- **4.4:** Bending the knees results in a **smaller** force compared to landing without bending, because it increases the time over which the dancer stops, reducing the required force.
**4.1:** The dancer's momentum upon landing is **250 kg·m/s** downward. **4.2:** Impulse is the change in momentum caused by a force over time. The net force acting on the dancer as she lands is **1250 N** upward. **4.4:** The force is **smaller than** the force calculated in 4.2 because bending the knees increases the time over which the dancer stops, reducing the required force during landing.

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