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Two identical balls A and B collide head on elastically. If velocities of A and B, before the collision are +0.5 m/s and -0.3 m/s respectively, then their velocities, after the collision, are respectively
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved. 1. Let the initial velocities be: Velocity of A before collision (uA) = +0.5 m/s Velocity of B before collision (uB) = -0.3 m/Read more
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved.
1. Let the initial velocities be:
Velocity of A before collision (uA) = +0.5 m/s
Velocity of B before collision (uB) = -0.3 m/s
2. For two equal masses in an elastic collision, the final velocities can be determined using the following formulas:
vA = (uA + uB) / 2 + (uA – uB) / 2
vB = (uA + uB) / 2 – (uA – uB) / 2
– First, we find the average velocity:
Average velocity (u_avg) = (uA + uB) / 2 = (0.5 – 0.3) / 2 = 0.1 / 2 = 0.05 m/s.
Now, we compute the change in velocities:
Change in velocity = uA – uB = 0.5 – (-0.3) = 0.5 + 0.3 = 0.8 m/s.
3. Using the equations above:
– Final velocity of A (vA):
vA = 0.05 + 0.4 = 0.45 m/s to be corrected depending on the direction.
– Final velocity of B (vB):
vB = 0.05 – 0.4 = -0.35 m/s to be corrected depending on the direction.
4. In a head-on collision, however, for two identical masses, their velocities will swap:
– After collision, ball A takes B’s initial velocity, and ball B takes A’s initial velocity.
– Thus, the final velocities will be:
– vA = -0.3 m/s (B’s initial velocity)
– vB = +0.5 m/s (A’s initial velocity)
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Does the ‘stopping potential’ in photoelectric emission depend upon (i) the intensity of the incident radiation in a photocell? (ii) the frequency of the incident radiation?
(i) The stopping potential does not depend on the intensity of the incident radiation, as intensity affects the number of photoelectrons, not their energy. (ii) The stopping potential depends on the frequency of the incident radiation because higher frequency photons impart more energy to the emitteRead more
(i) The stopping potential does not depend on the intensity of the incident radiation, as intensity affects the number of photoelectrons, not their energy.
(ii) The stopping potential depends on the frequency of the incident radiation because higher frequency photons impart more energy to the emitted electrons, requiring greater potential to stop them.
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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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