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What is the center of mass of a uniform semicircular ring of radius 𝑅 ?
The center of mass of a uniform semicircular ring lies at a distance 2𝑅/𝜋 along the axis of symmetry, derived using integration over the arc. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational motion. Give answer according to your underRead more
The center of mass of a uniform semicircular ring lies at a distance 2𝑅/𝜋 along the axis of symmetry, derived using integration over the arc.
This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational motion. Give answer according to your understanding.
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A ball moves in a frictionless inclined table without slipping. The work done by the table surface on the ball is
In the case of a frictionless inclined table, the work done by the table surface on the ball can be analyzed through the forces acting on the ball. Since there is no friction, the only force acting parallel to the surface is gravity. Gravity does not do work against the normal force of the table. ThRead more
In the case of a frictionless inclined table, the work done by the table surface on the ball can be analyzed through the forces acting on the ball.
Since there is no friction, the only force acting parallel to the surface is gravity. Gravity does not do work against the normal force of the table. The normal force acts perpendicular to the displacement of the ball.
Work done (W) is given by the formula:
W = F • d • cos(θ)
Where:
– F is the force
– d is the displacement
– θ is the angle between the force and displacement
This implies that the displacement and the force exerted are perpendicular to each other, that is, θ = 90 degrees. This gives cos(90°) = 0. Thus the work done by the table surface on the ball is:
W = F • d • 0 = 0
Final Answer:
The work done by the table surface on the ball is zero.
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A body of mass 10 kg initially at rest acquires velocity of 10 ms⁻¹. What is the work done?
To calculate the work done on the body, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. Step 1: Calculate the initial kinetic energy (K.E.₁) Since the body is initially at rest, its initial kinetic energy is: K.E.₁ = (1Read more
To calculate the work done on the body, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
Step 1: Calculate the initial kinetic energy (K.E.₁)
Since the body is initially at rest, its initial kinetic energy is:
K.E.₁ = (1/2) m v₁²
K.E.₁ = (1/2) × 10 kg × (0 m/s)²
K.E.₁ = 0 J
Step 2: Final kinetic energy K.E.₂
When the body has achieved a velocity of 10 m/s, the final kinetic energy is:
K.E.₂ = (1/2) m v₂²
K.E.₂ = (1/2) × 10 kg × (10 m/s)²
K.E.₂ = (1/2) × 10 × 100
K.E.₂ = 500 J
Step 3: Work done (W)
The work done equals the change in kinetic energy
W = K.E.₂ – K.E.₁
W = 500 J – 0 J
W = 500 J
Final Answer:
Work done is 500 J.
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A ball is dropped from height h on the ground where coefficient of restitution is e. After one balance the maximum height is
We can find the maximum height reached by a ball after it is dropped from height h and bounces on the ground with a coefficient of restitution e using the following analysis: 1. When the ball is dropped from height h, it gains kinetic energy just before hitting the ground. The potential energy at heRead more
We can find the maximum height reached by a ball after it is dropped from height h and bounces on the ground with a coefficient of restitution e using the following analysis:
1. When the ball is dropped from height h, it gains kinetic energy just before hitting the ground. The potential energy at height h is converted to kinetic energy:
Potential Energy (PE) = mgh
Kinetic Energy (KE) at the moment of impact = mgh
2. When it bounces, some energy is lost due to the coefficient of restitution e. The coefficient of restitution is defined as the ratio of the velocity after the bounce to the velocity before the bounce:
e = (velocity after bounce) / (velocity before bounce)
3. Velocity at the instant of hitting the ground (v) is determined by the relation:
v = √(2gh)
4. Just after the bounce, the velocity is given by,
velocity after bounce = e * v = e * √(2gh)
5. Maximum height (h’) attained just after the bounce can be calculated from the K.E. at the instant of bounce
K.E. after bounce = (1/2) m (e * √(2gh))²
This kinetic energy is turned back into potential energy at the maximum height:
PE at max height = mgh’
Thus, (1/2) m (e² * 2gh) = mgh’
Simplifying gives:
h’ = e²h
Final Answer:
The maximum height after the bounce is e²h.
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A particle of mass m having velocity v moving towards north collides with similar particle moving with same velocity towards east. The two particles stick together and move towards north east with a velocity
To find the velocity of two particles that collide and stick together, we can use the principles of conservation of momentum. Let: - Mass of each particle = m - Initial velocity of the first particle (moving north) = v - Initial velocity of the second particle (moving east) = v 1. Momentum Before CoRead more
To find the velocity of two particles that collide and stick together, we can use the principles of conservation of momentum.
Let:
– Mass of each particle = m
– Initial velocity of the first particle (moving north) = v
– Initial velocity of the second particle (moving east) = v
1. Momentum Before Collision:
– Momentum of the first particle (north): p₁ = m * v
– Momentum of the second particle (east): p₂ = m * v
2. Total Momentum Before Collision:
– The momentum vector of the first particle is (0, mv) (north direction).
– The momentum vector of the second particle is (mv, 0) (east direction).
– Therefore, the total momentum vector before the collision is:
P_initial = (mv, mv)
3. Momentum After Collision:
– Since the two particles stick together after the collision, the combined mass is 2m.
– Let the velocity of the combined mass after the collision be V, and its direction will be towards the northeast.
4. Using Pythagoras’ Theorem:
– The magnitude of the momentum vector after the collision can be found using:
P_final = √[(mv)² + (mv)²] = √[2(mv)²] = mv√2
5. Calculating the Final Velocity:
– The total momentum after the collision is equal to the momentum before the collision:
P_final = 2m * V
– Setting them equal:
mv√2 = 2m * V
– Canceling m from both sides:
v√2 = 2V
– Solving for V:
V = (v√2) / 2 = v / √2
Final Answer:
See lessThe velocity of the combined mass after the collision is v/√2.