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Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9?
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Body A of mass 4 m moving with speed u collides with another body B of...
Given data:
- Mass of body A, m_A = 4m
- Mass of body B, m_B = 2m
- Initial velocity of body A, u_A = u (moving)
- Initial velocity of body B, u_B = 0 (at rest)
- Collision type: Elastic
Key points:
1. In an elastic collision, both momentum and kinetic energy are conserved.
2. The fraction of energy lost by body A can be calculated using the equation:
Fraction of energy lost = (Initial kinetic energy - Final kinetic energy) / Initial kinetic energy
Solution:
Step 1: Calculate initial momentum:
- The initial momentum of body A is given by:
P_Ai = m_A * u_A
P_Ai = 4m * u
- The initial momentum of body B is zero as it is at rest:
P_Bi = m_B * u_B
P_Bi = 0
Step 2: Calculate final velocities:
- Since the collision is elastic, both momentum and kinetic energy are conserved.
- Applying momentum conservation:
P_Ai + P_Bi = P_Af + P_Bf
4m * u + 0 = 4m * v_Af + 2m * v_Bf
4mu = 4mv_Af + 2mv_Bf (Equation 1)
- Applying kinetic energy conservation:
(1/2) * m_A * u_A^2 + (1/2) * m_B * u_B^2 = (1/2) * m_A * v_Af^2 + (1/2) * m_B * v_Bf^2
(1/2) * 4m * u^2 + 0 = (1/2) * 4m * v_Af^2 + (1/2) * 2m * v_Bf^2
2mu^2 = 2mv_Af^2 + 2mv_Bf^2 (Equation 2)
Step 3: Solve the equations:
- From Equation 1, we can rewrite it as:
4mu = 4mv_Af + 2mv_Bf
2mu = 2mv_Af + mv_Bf
- Substituting the value of 2mu from Equation 2:
2mv_Af + 2mv_Bf = 2mv_Af + mv_Bf
mv_Bf = 0
- Since the final velocity of body B is zero, the final velocity of body A can be calculated as:
2mu = 2mv_Af
v_Af = u
Step 4: Calculate the fraction of energy lost:
- The initial kinetic energy of body A is given by:
KE_Ai = (1/2) * m_A * u_A^2
KE_Ai = (1/2) * 4m * u^2
KE_Ai = 2mu^2
- The final kinetic energy of body A is given by:
KE_Af = (1/2) * m_A
- Mass of body A, m_A = 4m
- Mass of body B, m_B = 2m
- Initial velocity of body A, u_A = u (moving)
- Initial velocity of body B, u_B = 0 (at rest)
- Collision type: Elastic
Key points:
1. In an elastic collision, both momentum and kinetic energy are conserved.
2. The fraction of energy lost by body A can be calculated using the equation:
Fraction of energy lost = (Initial kinetic energy - Final kinetic energy) / Initial kinetic energy
Solution:
Step 1: Calculate initial momentum:
- The initial momentum of body A is given by:
P_Ai = m_A * u_A
P_Ai = 4m * u
- The initial momentum of body B is zero as it is at rest:
P_Bi = m_B * u_B
P_Bi = 0
Step 2: Calculate final velocities:
- Since the collision is elastic, both momentum and kinetic energy are conserved.
- Applying momentum conservation:
P_Ai + P_Bi = P_Af + P_Bf
4m * u + 0 = 4m * v_Af + 2m * v_Bf
4mu = 4mv_Af + 2mv_Bf (Equation 1)
- Applying kinetic energy conservation:
(1/2) * m_A * u_A^2 + (1/2) * m_B * u_B^2 = (1/2) * m_A * v_Af^2 + (1/2) * m_B * v_Bf^2
(1/2) * 4m * u^2 + 0 = (1/2) * 4m * v_Af^2 + (1/2) * 2m * v_Bf^2
2mu^2 = 2mv_Af^2 + 2mv_Bf^2 (Equation 2)
Step 3: Solve the equations:
- From Equation 1, we can rewrite it as:
4mu = 4mv_Af + 2mv_Bf
2mu = 2mv_Af + mv_Bf
- Substituting the value of 2mu from Equation 2:
2mv_Af + 2mv_Bf = 2mv_Af + mv_Bf
mv_Bf = 0
- Since the final velocity of body B is zero, the final velocity of body A can be calculated as:
2mu = 2mv_Af
v_Af = u
Step 4: Calculate the fraction of energy lost:
- The initial kinetic energy of body A is given by:
KE_Ai = (1/2) * m_A * u_A^2
KE_Ai = (1/2) * 4m * u^2
KE_Ai = 2mu^2
- The final kinetic energy of body A is given by:
KE_Af = (1/2) * m_A
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Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9?
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Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9?.
Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9?.
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Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9?, a detailed solution for Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9? has been provided alongside types of Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : and explain a)1/9 b)8/9 c)4/9 d)5/9? theory, EduRev gives you an
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