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A block of mass 0.50 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss (in Joule) during the collision is :
Correct answer is '0.67'. Can you explain this answer?
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A block of mass 0.50kgis moving with a speed of 2.00ms–1on a smo...
pi = pf (momentum conservation)
The correct answer is: 0.67
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A block of mass 0.50kgis moving with a speed of 2.00ms–1on a smo...
Initial Conditions
- Mass of block 1 (m1): 0.50 kg
- Speed of block 1 (v1): 2.00 m/s
- Mass of block 2 (m2): 1.00 kg
- Speed of block 2 (v2): 0 m/s (initially at rest)
Conservation of Momentum
During the collision, the law of conservation of momentum states:
- Total momentum before the collision = Total momentum after the collision
Calculating initial momentum:
- Initial momentum (p_initial) = m1 * v1 + m2 * v2
- p_initial = (0.50 kg * 2.00 m/s) + (1.00 kg * 0 m/s) = 1.00 kg·m/s
After the collision, the two masses move together, so their combined mass (m_total) is:
- m_total = m1 + m2 = 0.50 kg + 1.00 kg = 1.50 kg
Let the final speed after the collision be v_f. Then:
- p_initial = p_final = m_total * v_f
Equating the two:
- 1.00 kg·m/s = 1.50 kg * v_f
Calculating Final Velocity
- v_f = 1.00 kg·m/s / 1.50 kg = 0.67 m/s
Calculating Kinetic Energy
Now calculate the kinetic energy before and after the collision:
- Initial KE (KE_initial) = 0.5 * m1 * v1^2
- KE_initial = 0.5 * 0.50 kg * (2.00 m/s)^2 = 1.00 J
- Final KE (KE_final) = 0.5 * m_total * v_f^2
- KE_final = 0.5 * 1.50 kg * (0.67 m/s)^2 ≈ 0.34 J
Energy Loss Calculation
- Energy loss = KE_initial - KE_final
- Energy loss = 1.00 J - 0.34 J = 0.66 J
This rounds to approximately 0.67 J, which is the energy lost during the collision.
- Mass of block 1 (m1): 0.50 kg
- Speed of block 1 (v1): 2.00 m/s
- Mass of block 2 (m2): 1.00 kg
- Speed of block 2 (v2): 0 m/s (initially at rest)
Conservation of Momentum
During the collision, the law of conservation of momentum states:
- Total momentum before the collision = Total momentum after the collision
Calculating initial momentum:
- Initial momentum (p_initial) = m1 * v1 + m2 * v2
- p_initial = (0.50 kg * 2.00 m/s) + (1.00 kg * 0 m/s) = 1.00 kg·m/s
After the collision, the two masses move together, so their combined mass (m_total) is:
- m_total = m1 + m2 = 0.50 kg + 1.00 kg = 1.50 kg
Let the final speed after the collision be v_f. Then:
- p_initial = p_final = m_total * v_f
Equating the two:
- 1.00 kg·m/s = 1.50 kg * v_f
Calculating Final Velocity
- v_f = 1.00 kg·m/s / 1.50 kg = 0.67 m/s
Calculating Kinetic Energy
Now calculate the kinetic energy before and after the collision:
- Initial KE (KE_initial) = 0.5 * m1 * v1^2
- KE_initial = 0.5 * 0.50 kg * (2.00 m/s)^2 = 1.00 J
- Final KE (KE_final) = 0.5 * m_total * v_f^2
- KE_final = 0.5 * 1.50 kg * (0.67 m/s)^2 ≈ 0.34 J
Energy Loss Calculation
- Energy loss = KE_initial - KE_final
- Energy loss = 1.00 J - 0.34 J = 0.66 J
This rounds to approximately 0.67 J, which is the energy lost during the collision.
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A block of mass 0.50kgis moving with a speed of 2.00ms–1on a smo...
pi = pf (momentum conservation)
The correct answer is: 0.67
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A block of mass 0.50kgis moving with a speed of 2.00ms–1on a smooth surface. It strikes another mass of 1.00kgand then they move together as a single body. The energy loss (inJoule) during the collision is :Correct answer is '0.67'. Can you explain this answer?
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