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A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction.
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A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a st...
Let v be the velocity of the girl on the cart as the cart starts moving
the total momenta of the girl and cart before the interaction
=60kg*6 m/s + 6kg*0m/s
=360kg m/s
total momenta after interaction
=(60+6)kg * v m/s
=66 v kg m/s
According to the law of conservation of momentum,the total momentum is conserved during the interaction
that is,
66v=360
v=360/66
v=5.45m/s
the girl on cart would move with a velocity of 5.45m/s.
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A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a st...
Problem Statement:
A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels. The mass of the cart is 6kg. What is her velocity as the cart starts moving? Assume that no external force is acting in the horizontal direction.
Solution:
Step 1: Analyzing the initial state
Before the girl jumps onto the cart, she has a horizontal velocity of 6 m/s. The cart is stationary, so its initial velocity is 0 m/s.
Given:
- Mass of the girl (m1) = 60 kg
- Mass of the cart (m2) = 6 kg
- Initial velocity of the girl (u1) = 6 m/s
- Initial velocity of the cart (u2) = 0 m/s
Step 2: Applying the law of conservation of momentum
According to the law of conservation of momentum, the total momentum before the jump is equal to the total momentum after the jump.
Total momentum before the jump = Total momentum after the jump
Momentum (p) = mass (m) × velocity (v)
(m1 × u1) + (m2 × u2) = (m1 × v1) + (m2 × v2)
Plugging in the given values:
(60 kg × 6 m/s) + (6 kg × 0 m/s) = (60 kg × v1) + (6 kg × v2)
360 kg·m/s = 60 kg·v1 + 6 kg·v2
Step 3: Finding the velocity of the girl and the cart
Since the cart is initially at rest (u2 = 0 m/s), the equation simplifies to:
60 kg·v1 = 360 kg·m/s
v1 = 360 kg·m/s / 60 kg
v1 = 6 m/s
Therefore, the velocity of the girl as the cart starts moving is 6 m/s.
To find the velocity of the cart (v2), we can substitute the value of v1 back into the equation:
360 kg·m/s = 60 kg·(6 m/s) + 6 kg·v2
360 kg·m/s = 360 kg·m/s + 6 kg·v2
6 kg·v2 = 0 kg·m/s
v2 = 0 m/s
Therefore, the velocity of the cart is 0 m/s.
Conclusion:
When the girl jumps onto the stationary cart with frictionless wheels, her velocity remains the same at 6 m/s. The cart starts moving with a velocity of 0 m/s. The conservation of momentum principle ensures that the total momentum before and after the jump remains constant.
A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels. The mass of the cart is 6kg. What is her velocity as the cart starts moving? Assume that no external force is acting in the horizontal direction.
Solution:
Step 1: Analyzing the initial state
Before the girl jumps onto the cart, she has a horizontal velocity of 6 m/s. The cart is stationary, so its initial velocity is 0 m/s.
Given:
- Mass of the girl (m1) = 60 kg
- Mass of the cart (m2) = 6 kg
- Initial velocity of the girl (u1) = 6 m/s
- Initial velocity of the cart (u2) = 0 m/s
Step 2: Applying the law of conservation of momentum
According to the law of conservation of momentum, the total momentum before the jump is equal to the total momentum after the jump.
Total momentum before the jump = Total momentum after the jump
Momentum (p) = mass (m) × velocity (v)
(m1 × u1) + (m2 × u2) = (m1 × v1) + (m2 × v2)
Plugging in the given values:
(60 kg × 6 m/s) + (6 kg × 0 m/s) = (60 kg × v1) + (6 kg × v2)
360 kg·m/s = 60 kg·v1 + 6 kg·v2
Step 3: Finding the velocity of the girl and the cart
Since the cart is initially at rest (u2 = 0 m/s), the equation simplifies to:
60 kg·v1 = 360 kg·m/s
v1 = 360 kg·m/s / 60 kg
v1 = 6 m/s
Therefore, the velocity of the girl as the cart starts moving is 6 m/s.
To find the velocity of the cart (v2), we can substitute the value of v1 back into the equation:
360 kg·m/s = 60 kg·(6 m/s) + 6 kg·v2
360 kg·m/s = 360 kg·m/s + 6 kg·v2
6 kg·v2 = 0 kg·m/s
v2 = 0 m/s
Therefore, the velocity of the cart is 0 m/s.
Conclusion:
When the girl jumps onto the stationary cart with frictionless wheels, her velocity remains the same at 6 m/s. The cart starts moving with a velocity of 0 m/s. The conservation of momentum principle ensures that the total momentum before and after the jump remains constant.
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A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction.
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A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction. for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction. covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction..
A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction. for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction. covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.The mass of the cart is 6kg.What is her velocity as the cart starts moving?Assume that no external force is acting in horizontal direction..
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