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.
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.
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%.
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this. The largerRead more
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this.
The larger disc has a larger area and mass, so it is initially centered at the origin. The smaller disc that is removed is also centered at the same point. The area and mass of the larger disc are proportionally greater than those of the smaller disc. This difference in mass distribution affects the position of the new center of mass after the smaller disc is removed.
As we analyze the situation, we can infer that the removal of the smaller disc will shift the center of mass toward the larger disc. Applying the principles of mass distribution and balance, we find that the center of mass of the remaining shape is located at a distance of αR from the center of the larger disc. After calculations, we conclude that the value of α is one-third which indicates the relative position of the center of mass in the modified structure.
One rod of 1.4 m length has the masses of 0.3 kg and 0.7 kg at both its ends, and its energy is determined while it is in rotational motion with a given rotational point. Let us determine this point for that given rod with respect to which it can have its rotational energy minimized if this rod is sRead more
One rod of 1.4 m length has the masses of 0.3 kg and 0.7 kg at both its ends, and its energy is determined while it is in rotational motion with a given rotational point. Let us determine this point for that given rod with respect to which it can have its rotational energy minimized if this rod is set in rotation at this particular point.
In this case, to find the point of optimal value, one must consider the mass distribution along the rod. In this scenario, the center of mass is of utmost importance, as it denotes the balance point of the system. If the rod rotates about its center of mass, the distances of each mass from that point determine the rotational energy.
The distances of the masses from the point of rotation can directly affect the moment of inertia. It would be perfect to have the axis of rotation closer to the larger mass if it is intended to reduce the rotational energy. Thus, in this configuration, the rotational energy is minimized at a distance of 0.98 meters from the 0.3 kg mass. This setup will make the system function very efficiently, bringing the energy necessary for rotation to a minimum.
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.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
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°.
Click for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
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%.
Click here for more :
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
A circular disc of radius R is removed from a bigger circular disc of radius 2R, such that the circumferences of the disc coincide. The centre of mass of the new disc is αR from the centre of the bigger disc. The value of α is
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this. The largerRead more
For finding the value of α as in the extraction of a small disc from the larger disc we take the arrangement, we assume we have large disc with a radius of size 2R and small disc of radius R. After we take out small disc, now we want to know the location of new centre of mass due to this.
The larger disc has a larger area and mass, so it is initially centered at the origin. The smaller disc that is removed is also centered at the same point. The area and mass of the larger disc are proportionally greater than those of the smaller disc. This difference in mass distribution affects the position of the new center of mass after the smaller disc is removed.
As we analyze the situation, we can infer that the removal of the smaller disc will shift the center of mass toward the larger disc. Applying the principles of mass distribution and balance, we find that the center of mass of the remaining shape is located at a distance of αR from the center of the larger disc. After calculations, we conclude that the value of α is one-third which indicates the relative position of the center of mass in the modified structure.
See More:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
A rod of length 1.4 m and negligible mass has two masses of 0.3 kg and 0.7 kg tied to its two ends. Find the location of the point on this rod, where the rotational energy is minimum, when the rod is rotated about the point.
One rod of 1.4 m length has the masses of 0.3 kg and 0.7 kg at both its ends, and its energy is determined while it is in rotational motion with a given rotational point. Let us determine this point for that given rod with respect to which it can have its rotational energy minimized if this rod is sRead more
One rod of 1.4 m length has the masses of 0.3 kg and 0.7 kg at both its ends, and its energy is determined while it is in rotational motion with a given rotational point. Let us determine this point for that given rod with respect to which it can have its rotational energy minimized if this rod is set in rotation at this particular point.
In this case, to find the point of optimal value, one must consider the mass distribution along the rod. In this scenario, the center of mass is of utmost importance, as it denotes the balance point of the system. If the rod rotates about its center of mass, the distances of each mass from that point determine the rotational energy.
The distances of the masses from the point of rotation can directly affect the moment of inertia. It would be perfect to have the axis of rotation closer to the larger mass if it is intended to reduce the rotational energy. Thus, in this configuration, the rotational energy is minimized at a distance of 0.98 meters from the 0.3 kg mass. This setup will make the system function very efficiently, bringing the energy necessary for rotation to a minimum.
Click here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/