leiss, work QUESTION 4 (Start on a new pago.) Dancers have to leam many skills, Including how to land correctly. A dancer of mass 50 kg leaps Tnto the air and lands feet first on the ground. She lands on the ground with a velocity of \( 5 \mathrm{~m} \cdot \mathrm{~s}^{-1} \). As she lands, she bends her knees and comes to a complete stop in 0,2 seconds: \( 4.1 \quad \) Calculate the momentum with which the dancer reaches the ground. \( 4.2 \quad \) Define the term impulse of a force. \( 4.3 \quad \) Calculate the magnitude of the net force acting on the dancer as she lands. Assume that the dancer performs the same jump as before but lands without bending her knees. \( 4.4 \quad \) Will the force. now be GREATER THAN, SMALLER THAN or EQUAL. TO the force calculated in QUESTION 4.3 ? 4.5

Certainly! Let's address each part of **QUESTION 4** step by step. --- ### **Question 4.1** **Calculate the momentum with which the dancer reaches the ground.** **Solution:** 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) Plugging in the values: \[ p = 50 \, \text{kg} \times 5 \, \text{m/s} = 250 \, \text{kg} \cdot \text{m/s} \] **Answer:** The momentum with which the dancer reaches the ground is **250 kg·m/s**. --- ### **Question 4.2** **Define the term impulse of a force.** **Answer:** **Impulse** is a measure of the change in momentum of an object when a force is applied over a specific period of time. It is calculated using the formula: \[ \text{Impulse} (J) = F \times \Delta t \] where: - \( F \) is the average force applied - \( \Delta t \) is the time interval during which the force is applied Impulse has both magnitude and direction, making it a vector quantity. It effectively quantifies how a force affects the motion of an object. --- ### **Question 4.3** **Calculate the magnitude of the net force acting on the dancer as she lands.** **Solution:** To find the net force (\( F \)) acting on the dancer during landing, we can use the impulse-momentum theorem: \[ J = \Delta p = F \times \Delta t \] From **Question 4.1**, we know: \[ \Delta p = 250 \, \text{kg} \cdot \text{m/s} \] Given: \[ \Delta t = 0.2 \, \text{seconds} \] Rearranging the formula to solve for \( F \): \[ F = \frac{\Delta p}{\Delta t} = \frac{250 \, \text{kg} \cdot \text{m/s}}{0.2 \, \text{s}} = 1250 \, \text{N} \] **Answer:** The magnitude of the net force acting on the dancer as she lands is **1250 Newtons**. --- ### **Question 4.4** **Assume that the dancer performs the same jump as before but lands without bending her knees. Will the force now be GREATER THAN, SMALLER THAN, or EQUAL TO the force calculated in QUESTION 4.3?** **Answer:** **Greater Than.** **Explanation:** When the dancer lands **without bending her knees**, the time interval (\( \Delta t \)) over which she comes to a stop is **shorter** than when she bends her knees. Since force is inversely proportional to the time interval (\( F = \frac{\Delta p}{\Delta t} \)), a shorter \( \Delta t \) results in a **larger** force. Therefore, the force exerted on her body when landing without bending her knees will be **greater than** the 1250 Newtons calculated in Question 4.3. --- If you have **Question 4.5** or any further queries, feel free to provide the details, and I'd be happy to help!