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A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball is
- a)1 N
- b)10 N
- c)100 N
- d)0.1 N
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A ball weighing 0.01kg hits a hard surface vertically with a speed of ...
Since impulse I = F x Δt
And also Impulse I = Δp (i.e. change in linear momentum)
Change in momentum
= 0.01 x 5 – (-5)
= 0.10 N s
Impulse = 0.10 N s
Therefore 0.10 = F x 0.01
Or F = 10 N
Most Upvoted Answer
A ball weighing 0.01kg hits a hard surface vertically with a speed of ...
Given:
Mass of the ball, m = 0.01 kg
Initial velocity, u = 5 m/s
Final velocity, v = -5 m/s (since the ball rebounds with the same speed in the opposite direction)
Time taken, t = 0.01 s
To find: Average force exerted by the surface, F
Formula used:
Average force, F = (change in momentum) / time taken
Change in momentum, Δp = m(v-u)
Calculation:
Δp = m(v-u)
Δp = 0.01 kg x (-5 m/s - 5 m/s)
Δp = -0.2 kg m/s (negative sign indicates the change in direction of momentum)
F = Δp / t
F = -0.2 kg m/s / 0.01 s
F = -20 N (negative sign indicates the force exerted by the surface is in the opposite direction of motion of the ball)
But we need to find the magnitude of the force, so we take the absolute value of the force:
|F| = 20 N
Therefore, the average force exerted by the surface on the ball is 20 N, which is closest to option (b) 10 N.
Mass of the ball, m = 0.01 kg
Initial velocity, u = 5 m/s
Final velocity, v = -5 m/s (since the ball rebounds with the same speed in the opposite direction)
Time taken, t = 0.01 s
To find: Average force exerted by the surface, F
Formula used:
Average force, F = (change in momentum) / time taken
Change in momentum, Δp = m(v-u)
Calculation:
Δp = m(v-u)
Δp = 0.01 kg x (-5 m/s - 5 m/s)
Δp = -0.2 kg m/s (negative sign indicates the change in direction of momentum)
F = Δp / t
F = -0.2 kg m/s / 0.01 s
F = -20 N (negative sign indicates the force exerted by the surface is in the opposite direction of motion of the ball)
But we need to find the magnitude of the force, so we take the absolute value of the force:
|F| = 20 N
Therefore, the average force exerted by the surface on the ball is 20 N, which is closest to option (b) 10 N.
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A ball weighing 0.01kg hits a hard surface vertically with a speed of ...
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A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer?
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A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer?.
A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball weighing 0.01kg hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball isa)1 Nb)10 Nc)100 Nd)0.1 NCorrect answer is option 'B'. Can you explain this answer?.
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