Find the impulse acting on a 5 kg object moving with initial velocity 30 m/s, if the impulse causes the kinetic energy to double.
Find the impulse acting on a 5 kg object moving with initial velocity 30 m/s, if the impulse causes the kinetic energy to double.
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![**Problem Statement:**
Find the impulse acting on a 5 kg object moving with an initial velocity of 30 m/s, if the impulse causes the kinetic energy to double.
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**Solution Explanation:**
1. **Initial Kinetic Energy (KE₁):**
\[
KE₁ = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \, \text{kg} \times (30 \, \text{m/s})^2 = 2250 \, \text{J}
\]
2. **Final Kinetic Energy (KE₂):**
Since the kinetic energy doubles, \( KE₂ = 2 \times KE₁ = 4500 \, \text{J} \).
3. **Final Velocity (v₂):**
\[
KE₂ = \frac{1}{2}mv_2^2 \Rightarrow 4500 \, \text{J} = \frac{1}{2} \times 5 \, \text{kg} \times v_2^2
\]
Solving for \( v_2 \):
\[
4500 = 2.5 \times v_2^2 \Rightarrow v_2^2 = 1800 \Rightarrow v_2 = \sqrt{1800} \approx 42.43 \, \text{m/s}
\]
4. **Impulse (J):**
\[
J = \Delta p = m(v_2 - v_1) = 5 \times (42.43 - 30) = 5 \times 12.43 = 62.15 \, \text{Ns}
\]
The impulse required to double the kinetic energy of the object is \( 62.15 \, \text{Ns} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3738e16a-10fd-4378-b4a3-b8733d2d5bb5%2F4739ab1e-9284-41e7-8c15-382fce6bb9f0%2Fgcry2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the impulse acting on a 5 kg object moving with an initial velocity of 30 m/s, if the impulse causes the kinetic energy to double.
---
**Solution Explanation:**
1. **Initial Kinetic Energy (KE₁):**
\[
KE₁ = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \, \text{kg} \times (30 \, \text{m/s})^2 = 2250 \, \text{J}
\]
2. **Final Kinetic Energy (KE₂):**
Since the kinetic energy doubles, \( KE₂ = 2 \times KE₁ = 4500 \, \text{J} \).
3. **Final Velocity (v₂):**
\[
KE₂ = \frac{1}{2}mv_2^2 \Rightarrow 4500 \, \text{J} = \frac{1}{2} \times 5 \, \text{kg} \times v_2^2
\]
Solving for \( v_2 \):
\[
4500 = 2.5 \times v_2^2 \Rightarrow v_2^2 = 1800 \Rightarrow v_2 = \sqrt{1800} \approx 42.43 \, \text{m/s}
\]
4. **Impulse (J):**
\[
J = \Delta p = m(v_2 - v_1) = 5 \times (42.43 - 30) = 5 \times 12.43 = 62.15 \, \text{Ns}
\]
The impulse required to double the kinetic energy of the object is \( 62.15 \, \text{Ns} \).
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