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A particle of mass m moving with velocity v collides with a stationary particle of mass 2 m. After collision, the speed of the combined particle is
Using the principle of conservation of momentum, we can calculate the speed of the combined particle after the collision. 1. Let the initial momentum of the system be: - The momentum of the first particle having mass m and velocity v is m * v. - The momentum of the second particle having mass 2m andRead more
Using the principle of conservation of momentum, we can calculate the speed of the combined particle after the collision.
1. Let the initial momentum of the system be:
– The momentum of the first particle having mass m and velocity v is m * v.
– The momentum of the second particle having mass 2m and at rest is 2m * 0 = 0.
Total initial momentum = mv + 0 = mv.
2. As the collision takes place, the two particles merge together and a total mass as follows:
Total mass = m + 2m = 3m.
3. Supposing the speed of the merged particle after the collision as V. According to the law of conservation of momentum
Initial momentum = Final momentum
mv = (3m) * V.
4. Solving for V:
V = mv / (3m) = v / 3.
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The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?
The maximum kinetic energy of photoelectrons is given by K.E. = e × V, where e is the electron charge (1.6 × 10⁻¹⁹ C) and V is the stopping potential. Thus, K.E. = 1.6 × 10⁻¹⁹ × 1.5 = 2.4 × 10⁻¹⁹ J. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-11/
The maximum kinetic energy of photoelectrons is given by K.E. = e × V, where e is the electron charge (1.6 × 10⁻¹⁹ C) and V is the stopping potential. Thus, K.E. = 1.6 × 10⁻¹⁹ × 1.5 = 2.4 × 10⁻¹⁹ J.
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A body moves a distance 0f 10 m along a straight line under the action of a 5 N force. If the work done is 25 J, then angle between the force and direction of motion of the body is
We can use the work done formula to find the angle between the force and the direction of motion: Work done (W) = Force (F) × Distance (d) × cos(θ), where θ is the angle between the force and the direction of motion. 1. Given values: - Work done (W) = 25 J, - Force (F) = 5 N, - Distance (d) = 10 m.Read more
We can use the work done formula to find the angle between the force and the direction of motion:
Work done (W) = Force (F) × Distance (d) × cos(θ),
where θ is the angle between the force and the direction of motion.
1. Given values:
– Work done (W) = 25 J,
– Force (F) = 5 N,
– Distance (d) = 10 m.
2. Substituting the given values into the formula:
25 J = 5 N × 10 m × cos(θ).
3. Simplifying the equation:
25 = 50 × cos(θ).
4. Dividing both sides by 50:
cos(θ) = 25 / 50 = 0.5.
5. Now, finding the angle θ:
θ = cos⁻¹(0.5) = 60°.
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If momentum decreases by 20 %, K.E. will decrease by
To find the percentage decrease in kinetic energy (K.E.) when momentum decreases by 20%, we have the following relations: 1. Momentum (p) is defined as: p = m * v where m is the mass and v is the velocity. 2. Kinetic energy (K.E.) is defined as: K.E. = (1/2) * m * v². 3. If momentum decreases by 20%Read more
To find the percentage decrease in kinetic energy (K.E.) when momentum decreases by 20%, we have the following relations:
1. Momentum (p) is defined as:
p = m * v
where m is the mass and v is the velocity.
2. Kinetic energy (K.E.) is defined as:
K.E. = (1/2) * m * v².
3. If momentum decreases by 20%, then the new momentum is given by p’.
p’ = p – 0.2p = 0.8p.
4. As momentum is directly proportional to mass and velocity, we can express the new velocity, v’, in terms of the original velocity, v, as:
p’ = m * v’ = 0.8(m * v).
5. Divide both sides by m
v’ = 0.8v.
6. We can now compute the new kinetic energy, K.E.’:
K.E.’ = (1/2) * m * (v’)²
= (1/2) * m * (0.8v)²
= (1/2) * m * 0.64v²
= 0.64 * (1/2) * m * v²
= 0.64 * K.E.
7. The percentage decrease in kinetic energy is:
Percentage decrease = [(K.E. – K.E.’) / K.E.] * 100
= [(K.E. – 0.64K.E.) / K.E.] * 100
= (0.36K.E. / K.E.) * 100 = 36%.
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Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (1) red light, and (ii) blue light?
(1) Electrons will not be emitted with red light because its frequency is lower than green and yellow light, providing insufficient energy. (2) Electrons will be emitted with blue light as its frequency is higher than green light, exceeding the threshold frequency. For more visit here: https://www.tRead more
(1) Electrons will not be emitted with red light because its frequency is lower than green and yellow light, providing insufficient energy.
(2) Electrons will be emitted with blue light as its frequency is higher than green light, exceeding the threshold frequency.
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