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A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/are
- a)6 kg only
- b)1 kg, 9 kg
- c)1 kg only
- d)1 kg, 6 kg
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfec...
Conservation of momentum gives m1u1 + m2u2 = m1v1 + m2v2
m1=3kg, u1=4m/s, m2=m kg, u2=0
Substituting in the above equation,
3×4+m×0=3×v1+m×v2
Or, 12=3v1+mv2-------(1)
For perfectly-elastic impact,
Or, v2−v1=u1−u2=4m/s------- (2)
Conservation of kinetic energy gives
Or,
From equation (2), v1=v2−4
Substituting in equation (1),
12=3(v2−4)+mv2
Substituting in equation (3)
On solving, we get
m = 1,9
Therefore, possible values of m are 1 kg and 9 kg.
m1=3kg, u1=4m/s, m2=m kg, u2=0
Substituting in the above equation,
3×4+m×0=3×v1+m×v2
Or, 12=3v1+mv2-------(1)
For perfectly-elastic impact,
Or, v2−v1=u1−u2=4m/s------- (2)
Conservation of kinetic energy gives
Or,
From equation (2), v1=v2−4
Substituting in equation (1),
12=3(v2−4)+mv2
Substituting in equation (3)
On solving, we get
m = 1,9
Therefore, possible values of m are 1 kg and 9 kg.
Most Upvoted Answer
A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfec...
Let's analyze the problem step by step to determine the possible values of mass m.
Given:
Mass of the first ball (ball A) = 3 kg
Velocity of ball A before impact = 4 m/s
Kinetic energy of ball A after impact = 6 J
Step 1: Conservation of Momentum
In an elastic collision, both momentum and kinetic energy are conserved. Therefore, we can use the principle of conservation of momentum to analyze the collision.
Before the impact:
Momentum of ball A = mass of A × velocity of A = 3 kg × 4 m/s = 12 kg·m/s
After the impact:
Let the velocity of ball A after impact be v1 and the velocity of ball B (the stationary ball) after impact be v2.
Momentum of ball A after impact = mass of A × velocity of A after impact = 3 kg × v1
Momentum of ball B after impact = mass of B × velocity of B after impact = m × v2
Using the principle of conservation of momentum:
Momentum before impact = Momentum after impact
12 kg·m/s = 3 kg × v1 + m × v2 (Equation 1)
Step 2: Conservation of Kinetic Energy
Since the collision is perfectly elastic, the kinetic energy of the system is conserved. Therefore, we can use the principle of conservation of kinetic energy to analyze the collision.
Before the impact:
Kinetic energy of ball A = 0.5 × mass of A × (velocity of A)^2 = 0.5 × 3 kg × (4 m/s)^2 = 24 J
After the impact:
Kinetic energy of ball A = 0.5 × mass of A × (velocity of A after impact)^2 = 0.5 × 3 kg × (v1)^2 = 6 J
Kinetic energy of ball B = 0.5 × mass of B × (velocity of B after impact)^2 = 0.5 × m × (v2)^2
Using the principle of conservation of kinetic energy:
Kinetic energy before impact = Kinetic energy after impact
24 J = 6 J + 0.5 × m × (v2)^2 (Equation 2)
Step 3: Solving the Equations
We have two equations (Equation 1 and Equation 2) with two unknowns (v1 and v2). By solving these equations simultaneously, we can determine the possible values of mass m.
From Equation 1, we can express v2 in terms of v1:
v2 = (12 kg·m/s - 3 kg × v1) / m
Substituting this expression for v2 into Equation 2, we get:
24 J = 6 J + 0.5 × m × ((12 kg·m/s - 3 kg × v1) / m)^2
Simplifying the equation:
18 J = 0.5 × ((12 kg·m/s - 3 kg × v1) / m)^2
Further simplification:
36 = ((12 kg·m/s - 3 kg × v1) / m)^2
Taking the square root of both sides:
6 = (12 kg·m/s - 3 kg × v1) / m
Cross-multiplying
Given:
Mass of the first ball (ball A) = 3 kg
Velocity of ball A before impact = 4 m/s
Kinetic energy of ball A after impact = 6 J
Step 1: Conservation of Momentum
In an elastic collision, both momentum and kinetic energy are conserved. Therefore, we can use the principle of conservation of momentum to analyze the collision.
Before the impact:
Momentum of ball A = mass of A × velocity of A = 3 kg × 4 m/s = 12 kg·m/s
After the impact:
Let the velocity of ball A after impact be v1 and the velocity of ball B (the stationary ball) after impact be v2.
Momentum of ball A after impact = mass of A × velocity of A after impact = 3 kg × v1
Momentum of ball B after impact = mass of B × velocity of B after impact = m × v2
Using the principle of conservation of momentum:
Momentum before impact = Momentum after impact
12 kg·m/s = 3 kg × v1 + m × v2 (Equation 1)
Step 2: Conservation of Kinetic Energy
Since the collision is perfectly elastic, the kinetic energy of the system is conserved. Therefore, we can use the principle of conservation of kinetic energy to analyze the collision.
Before the impact:
Kinetic energy of ball A = 0.5 × mass of A × (velocity of A)^2 = 0.5 × 3 kg × (4 m/s)^2 = 24 J
After the impact:
Kinetic energy of ball A = 0.5 × mass of A × (velocity of A after impact)^2 = 0.5 × 3 kg × (v1)^2 = 6 J
Kinetic energy of ball B = 0.5 × mass of B × (velocity of B after impact)^2 = 0.5 × m × (v2)^2
Using the principle of conservation of kinetic energy:
Kinetic energy before impact = Kinetic energy after impact
24 J = 6 J + 0.5 × m × (v2)^2 (Equation 2)
Step 3: Solving the Equations
We have two equations (Equation 1 and Equation 2) with two unknowns (v1 and v2). By solving these equations simultaneously, we can determine the possible values of mass m.
From Equation 1, we can express v2 in terms of v1:
v2 = (12 kg·m/s - 3 kg × v1) / m
Substituting this expression for v2 into Equation 2, we get:
24 J = 6 J + 0.5 × m × ((12 kg·m/s - 3 kg × v1) / m)^2
Simplifying the equation:
18 J = 0.5 × ((12 kg·m/s - 3 kg × v1) / m)^2
Further simplification:
36 = ((12 kg·m/s - 3 kg × v1) / m)^2
Taking the square root of both sides:
6 = (12 kg·m/s - 3 kg × v1) / m
Cross-multiplying
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A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer?
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A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer?.
A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value (s) of m is/area)6 kg onlyb)1 kg, 9 kgc)1 kg onlyd)1 kg, 6 kgCorrect answer is option 'D'. Can you explain this answer?.
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