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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 during the collision is
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- a)0.16 J
- b)1.00 J
- c)0.67 J
- d)0.34 J
Correct answer is option 'C'. Can you explain this answer?
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To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of Momentum:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
m1 * v1 + m2 * v2 = (m1 + m2) * v
where m1 and m2 are the masses of the two objects, v1 and v2 are their initial velocities, and v is the final velocity of the combined system.
In this case, the initial momentum of the 0.50 kg block is given by:
m1 * v1 = 0.50 kg * 2.00 m/s = 1.00 kg·m/s
Let's assume the final velocity of the combined system is v_f. Using the conservation of momentum equation, we can solve for v_f:
1.00 kg·m/s + 1.00 kg * 0 m/s = (0.50 kg + 1.00 kg) * v_f
1.00 kg·m/s = 1.50 kg * v_f
v_f = 1.00 kg·m/s / 1.50 kg
v_f = 0.67 m/s
2. Conservation of Kinetic Energy:
The initial kinetic energy of the 0.50 kg block is given by:
KE1 = (1/2) * m1 * v1^2
KE1 = (1/2) * 0.50 kg * (2.00 m/s)^2
KE1 = 1.00 J
The final kinetic energy of the combined system can be calculated using the final velocity:
KE_f = (1/2) * (m1 + m2) * v_f^2
KE_f = (1/2) * (0.50 kg + 1.00 kg) * (0.67 m/s)^2
KE_f = 0.67 J
The energy loss during the collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy:
Energy loss = KE1 - KE_f
Energy loss = 1.00 J - 0.67 J
Energy loss = 0.33 J
Therefore, the correct answer is option C) 0.67 J.
1. Conservation of Momentum:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
m1 * v1 + m2 * v2 = (m1 + m2) * v
where m1 and m2 are the masses of the two objects, v1 and v2 are their initial velocities, and v is the final velocity of the combined system.
In this case, the initial momentum of the 0.50 kg block is given by:
m1 * v1 = 0.50 kg * 2.00 m/s = 1.00 kg·m/s
Let's assume the final velocity of the combined system is v_f. Using the conservation of momentum equation, we can solve for v_f:
1.00 kg·m/s + 1.00 kg * 0 m/s = (0.50 kg + 1.00 kg) * v_f
1.00 kg·m/s = 1.50 kg * v_f
v_f = 1.00 kg·m/s / 1.50 kg
v_f = 0.67 m/s
2. Conservation of Kinetic Energy:
The initial kinetic energy of the 0.50 kg block is given by:
KE1 = (1/2) * m1 * v1^2
KE1 = (1/2) * 0.50 kg * (2.00 m/s)^2
KE1 = 1.00 J
The final kinetic energy of the combined system can be calculated using the final velocity:
KE_f = (1/2) * (m1 + m2) * v_f^2
KE_f = (1/2) * (0.50 kg + 1.00 kg) * (0.67 m/s)^2
KE_f = 0.67 J
The energy loss during the collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy:
Energy loss = KE1 - KE_f
Energy loss = 1.00 J - 0.67 J
Energy loss = 0.33 J
Therefore, the correct answer is option C) 0.67 J.
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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is[AIEEE 2008]a)0.16 Jb)1.00 Jc)0.67 Jd)0.34 JCorrect answer is option 'C'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of mass 0.50 kg is moving with a speed of 2.00 ms-1on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is[AIEEE 2008]a)0.16 Jb)1.00 Jc)0.67 Jd)0.34 JCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 0.50 kg is moving with a speed of 2.00 ms-1on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is[AIEEE 2008]a)0.16 Jb)1.00 Jc)0.67 Jd)0.34 JCorrect answer is option 'C'. Can you explain this answer?.
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is[AIEEE 2008]a)0.16 Jb)1.00 Jc)0.67 Jd)0.34 JCorrect answer is option 'C'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of mass 0.50 kg is moving with a speed of 2.00 ms-1on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is[AIEEE 2008]a)0.16 Jb)1.00 Jc)0.67 Jd)0.34 JCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 0.50 kg is moving with a speed of 2.00 ms-1on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is[AIEEE 2008]a)0.16 Jb)1.00 Jc)0.67 Jd)0.34 JCorrect answer is option 'C'. Can you explain this answer?.
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