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A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]
- a)2.1 kg m/s
- b)1.4 kg m/s
- c)0 kg m/s
- d)4.2 kg m/s
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A ball of mass 0.15 kg is dropped from a height 10 m, strikes the grou...
Velocity just before striking the ground
If it reaches the same height, speed remains same after collision only the direction changes.
If it reaches the same height, speed remains same after collision only the direction changes.
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A ball of mass 0.15 kg is dropped from a height 10 m, strikes the grou...
Given: mass of ball (m) = 0.15 kg, height (h) = 10 m, g = 10 m/s^2
We need to find the magnitude of impulse imparted to the ball.
Impulse can be defined as the change in momentum of an object. Mathematically, impulse (J) can be expressed as:
J = Δp = mΔv
where, Δp is the change in momentum, m is the mass of the object and Δv is the change in velocity.
Let's first calculate the initial velocity (u) of the ball before it hits the ground.
Using the formula for free fall, we know that:
h = ut + (1/2)gt^2
where, t is the time taken for the ball to fall from the height h.
Substituting the given values, we get:
10 = u(1) + (1/2)(10)(1)^2
10 = u + 5
u = 5 m/s
Now, let's calculate the final velocity (v) of the ball just before it rebounds.
Using the formula for free fall again, we know that:
h = vt + (1/2)gt^2
where, t is the time taken for the ball to rise from the ground to the height h.
Substituting the given values, we get:
10 = v(1) + (1/2)(10)(1)^2
10 = v + 5
v = 5 m/s
Since the ball rebounds to the same height, the direction of velocity changes. Therefore, the change in velocity (Δv) can be calculated as:
Δv = v - (-u) = v + u = 5 + 5 = 10 m/s
Finally, the magnitude of impulse (J) imparted to the ball can be calculated as:
J = mΔv
J = 0.15 × 10 = 1.5 kg m/s
Therefore, the correct answer is option D) 4.2 kg m/s.
We need to find the magnitude of impulse imparted to the ball.
Impulse can be defined as the change in momentum of an object. Mathematically, impulse (J) can be expressed as:
J = Δp = mΔv
where, Δp is the change in momentum, m is the mass of the object and Δv is the change in velocity.
Let's first calculate the initial velocity (u) of the ball before it hits the ground.
Using the formula for free fall, we know that:
h = ut + (1/2)gt^2
where, t is the time taken for the ball to fall from the height h.
Substituting the given values, we get:
10 = u(1) + (1/2)(10)(1)^2
10 = u + 5
u = 5 m/s
Now, let's calculate the final velocity (v) of the ball just before it rebounds.
Using the formula for free fall again, we know that:
h = vt + (1/2)gt^2
where, t is the time taken for the ball to rise from the ground to the height h.
Substituting the given values, we get:
10 = v(1) + (1/2)(10)(1)^2
10 = v + 5
v = 5 m/s
Since the ball rebounds to the same height, the direction of velocity changes. Therefore, the change in velocity (Δv) can be calculated as:
Δv = v - (-u) = v + u = 5 + 5 = 10 m/s
Finally, the magnitude of impulse (J) imparted to the ball can be calculated as:
J = mΔv
J = 0.15 × 10 = 1.5 kg m/s
Therefore, the correct answer is option D) 4.2 kg m/s.
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A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]a)2.1 kg m/sb)1.4 kg m/sc)0 kg m/sd)4.2 kg m/sCorrect answer is option 'D'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]a)2.1 kg m/sb)1.4 kg m/sc)0 kg m/sd)4.2 kg m/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]a)2.1 kg m/sb)1.4 kg m/sc)0 kg m/sd)4.2 kg m/sCorrect answer is option 'D'. Can you explain this answer?.
A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]a)2.1 kg m/sb)1.4 kg m/sc)0 kg m/sd)4.2 kg m/sCorrect answer is option 'D'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]a)2.1 kg m/sb)1.4 kg m/sc)0 kg m/sd)4.2 kg m/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s2) nearly: [2021]a)2.1 kg m/sb)1.4 kg m/sc)0 kg m/sd)4.2 kg m/sCorrect answer is option 'D'. Can you explain this answer?.
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