An ice skater 'A' of total mass 65kg slides on a frictionless frozen lake ina straight line with velocity 2m/s?
while moving,A throws his helmet weight 5kg horizontally with a velocity of 4m/s in the direction perpendicular to his direction of motion .
(a) Find the magnitude of the resultant velocity of A after throwing the helmet.
(b) Another skater 'B' of mass 45kg sliding freely in the opposite direction on a nearby parallel course with velocity 1m/s catches the helmet thrown by A.after catching the helmet,find the new velocity of B in;
(1)the original direction of motion of B
(2)the direction perpendicular to the original direction of motion B.
(c) Calculate the total kinetic energy of the helmet and the skater B just before B catches the helmet.
(d)Calculate also the kinetic energy of the helmet and the skater B after B catches the helmet.
(e) Explain briefly why the two values calculated in (c) and (d) above are different.
(f) After some time the helmet falls off freely from the skater B. What will happen to the velocity of B? explain your answer.
Idk manz
(a) The magnitude of the resultant velocity of A after throwing the helmet can be found using the principle of conservation of momentum. The momentum of A before throwing the helmet is given by mass of A multiplied by velocity of A, which is 65 kg * 2 m/s = 130 kg*m/s. The momentum of the helmet is given by mass of the helmet multiplied by velocity of the helmet, which is 5 kg * 4 m/s = 20 kg*m/s. Since the system is isolated and there is no external force, the total momentum of A and the helmet before throwing will be equal to the total momentum after throwing. Therefore, the magnitude of the resultant velocity of A can be calculated by dividing the total momentum by the mass of A: 130 kg*m/s / 65 kg = 2 m/s.
(b)
(1) When skater B catches the helmet and there is no external force acting on the system, the total momentum of the system will be conserved. Since skater B catches the helmet in the same direction as his original motion, the velocity of skater B will increase. The new velocity of B can be calculated using the principle of conservation of momentum. The momentum of A is 130 kg*m/s, and the momentum of the helmet is 20 kg*m/s. So, the total momentum of the system is 130 kg*m/s + 20 kg*m/s = 150 kg*m/s. Dividing the total momentum by the mass of B, we get: 150 kg*m/s / 45 kg = 3.33 m/s.
(2) The direction perpendicular to the original direction of motion of B remains unchanged when skater B catches the helmet. Therefore, the new velocity of B in this direction will remain the same as before, which is 1 m/s.
(c) The total kinetic energy of the helmet and skater B just before B catches the helmet can be calculated by summing the individual kinetic energies of the helmet and B. The formula for kinetic energy is 1/2 * mass * velocity^2. The kinetic energy of the helmet is 1/2 * 5 kg * 4 m/s^2 = 40 J. The kinetic energy of skater B is 1/2 * 45 kg * 1 m/s^2 = 22.5 J. Summing these values, we get the total kinetic energy just before B catches the helmet: 40 J + 22.5 J = 62.5 J.
(d) After B catches the helmet, the velocities of both the helmet and B change. To calculate the kinetic energy after catching the helmet, we need to know the new velocities of both the helmet and B. Since the problem does not provide these values, we cannot calculate the kinetic energy after catching.
(e) The two values calculated in (c) and (d) above are different because the kinetic energy depends on the square of the velocity. When skater B catches the helmet, both velocities change, thus affecting the kinetic energy.
(f) When the helmet falls off freely from skater B, the velocity of skater B will remain unchanged, assuming there are no external forces acting on the system. The reason is that the falling off of the helmet does not exert any force on skater B, so there is no change in momentum, and hence no change in velocity.
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
(a) To find the magnitude of the resultant velocity of skater A after throwing the helmet, we can use the principle of conservation of momentum.
The initial momentum of skater A is given by:
P_initial = mass_A * velocity_A = 65 kg * 2 m/s = 130 kg·m/s
The momentum of the thrown helmet is given by:
P_helmet = mass_helmet * velocity_helmet = 5 kg * 4 m/s = 20 kg·m/s
Since the velocity of the helmet is perpendicular to the direction of motion of skater A, it does not affect the magnitude of A's velocity.
Therefore, the magnitude of the resultant velocity of skater A after throwing the helmet is still 2 m/s.
(b) After catching the helmet, the new velocity of skater B can be found by using the principle of conservation of momentum.
Let's analyze the velocities before and after the helmet is caught:
Before:
- Skater A has a velocity of 2 m/s in one direction.
- Skater B has a velocity of 1 m/s in the opposite direction.
After:
- Skater A still has a velocity of 2 m/s.
- Skater B catches the helmet, adding its momentum to his own.
Applying the principle of conservation of momentum:
P_before = P_after
(mass_A * velocity_A) + (mass_B * velocity_B) = (mass_A * new_velocity_A) + (mass_B * new_velocity_B)
Substituting the given values:
(65 kg * 2 m/s) + (45 kg * 1 m/s) = (65 kg * new_velocity_A) + (45 kg * new_velocity_B)
Simplifying the equation:
130 kg·m/s + 45 kg·m/s = 65 kg * new_velocity_A + 45 kg * new_velocity_B
175 kg·m/s = 65 kg * new_velocity_A + 45 kg * new_velocity_B
Now, we can solve for the new velocities.
(1) The original direction of motion of B:
The new_velocity_A = 2 m/s (remains unchanged)
175 kg·m/s = 65 kg * 2 m/s + 45 kg * new_velocity_B
175 kg·m/s - 65 kg * 2 m/s = 45 kg * new_velocity_B
175 kg·m/s - 130 kg·m/s = 45 kg * new_velocity_B
45 kg * new_velocity_B = 45 kg·m/s
new_velocity_B = 1 m/s
Therefore, the new velocity of skater B in the original direction of motion is 1 m/s.
(2) The direction perpendicular to the original direction of motion B:
In this direction, the velocity of skater B does not change because the momentum is conserved only in the direction of motion.
Therefore, the new velocity of skater B in the direction perpendicular to the original direction of motion is still 0 m/s.
(c) The total kinetic energy of the helmet and skater B just before B catches the helmet can be calculated as follows:
The kinetic energy (KE) is given by:
KE = 0.5 * mass * velocity^2
KE_before = 0.5 * (mass_helmet + mass_B) * velocity_B^2
= 0.5 * (5 kg + 45 kg) * 1 m/s^2
= 25 J
Therefore, the total kinetic energy of the helmet and skater B just before B catches the helmet is 25 Joules.
(d) After B catches the helmet, the kinetic energy can be calculated as follows:
KE_after = 0.5 * (mass_helmet + mass_B) * new_velocity_B^2
= 0.5 * (5 kg + 45 kg) * 1 m/s^2
= 25 J
Therefore, the kinetic energy of the helmet and skater B after B catches the helmet is still 25 Joules.
(e) The two values calculated in (c) and (d) are different because the principle of conservation of kinetic energy applies only if there are no external forces doing work on the system. In this case, skater B catching the helmet involves an external force, so some energy is transferred in the process, resulting in a change in kinetic energy.
(f) When the helmet falls off freely from skater B, the velocity of B will not change. This is because the helmet falling off does not result in any external force acting on skater B in the horizontal direction. The only force acting on B is the force due to gravity, which does not cause a change in the horizontal velocity of B.
To solve this problem, we will use the principles of conservation of linear momentum and conservation of kinetic energy. Let's break it down step by step:
(a) To find the magnitude of the resultant velocity of Skater A after throwing the helmet, we need to consider the conservation of linear momentum.
The initial momentum of Skater A and the helmet can be calculated by multiplying their respective masses by their initial velocities.
Initial momentum of Skater A = mass of Skater A * initial velocity of Skater A
= 65 kg * 2 m/s
The final momentum of Skater A after throwing the helmet will be equal to the initial momentum since there are no external forces acting on Skater A.
Final momentum of Skater A = mass of Skater A * final velocity of Skater A
Since momentum is a vector quantity, the final momentum of Skater A will have both magnitude and direction. To find the magnitude of the resultant velocity, we can use the Pythagorean theorem:
(final velocity of Skater A)^2 = (initial velocity of Skater A)^2 + (velocity of thrown helmet)^2
Plug in the values:
(final velocity of Skater A)^2 = (2 m/s)^2 + (4 m/s)^2
Solve for the magnitude of the resultant velocity of Skater A.
(b) After catching the helmet, Skater B interacts with the thrown helmet, resulting in a transfer of momentum. To find the new velocity of Skater B in different directions, we need to use the conservation of linear momentum.
(1) In the original direction of motion of Skater B:
Since Skater A and Skater B are sliding freely on a frictionless surface, the net external force acting on the system is zero. Therefore, the total linear momentum of the system before and after the interaction remains conserved.
Initial momentum of the system = Final momentum of the system
Let's denote the initial velocity of Skater B in the original direction as Vb1 and the final velocity as Vb1'.
Initial momentum of Skater B = mass of Skater B * initial velocity of Skater B
= 45 kg * 1 m/s
Final momentum of Skater B = mass of Skater B * final velocity of Skater B in the original direction
= 45 kg * Vb1'
Using the conservation of momentum, we can equate the two expressions to find Vb1'.
(2) In the direction perpendicular to the original direction of motion of Skater B:
Similar to the previous case, the total linear momentum of the system before and after the interaction remains conserved.
Initial momentum of the system = Final momentum of the system
Let's denote the final velocity of Skater B in the direction perpendicular to the original direction as Vb2'.
Initial momentum of Skater B = mass of Skater B * initial velocity of Skater B
= 45 kg * 1 m/s
Final momentum of Skater B = mass of Skater B * final velocity of Skater B in the direction perpendicular to the original direction
= 45 kg * Vb2'
Using the conservation of momentum, we can equate the two expressions to find Vb2'.
(c) To calculate the total kinetic energy of the helmet and Skater B just before B catches the helmet, we need to consider the conservation of kinetic energy.
Total initial kinetic energy = kinetic energy of Skater B + kinetic energy of the helmet
The kinetic energy of an object is given by the formula:
Kinetic energy = (1/2) * mass * velocity^2
Substitute the masses and velocities of Skater B and the helmet into the formula to calculate the initial kinetic energy.
(d) To calculate the kinetic energy of the helmet and Skater B after B catches the helmet, we again consider the conservation of kinetic energy.
Total final kinetic energy = kinetic energy of Skater B + kinetic energy of the helmet
Substitute the masses and velocities of Skater B and the helmet into the formula to calculate the final kinetic energy.
(e) The two values calculated in (c) and (d) may be different due to the transfer of kinetic energy from Skater A to Skater B and the helmet. When Skater B catches the helmet, there is a change in the distribution of kinetic energy within the system. Some of the initial kinetic energy of Skater B gets transferred to the helmet, resulting in a change in the total kinetic energy of the system.
(f) When the helmet falls off freely from Skater B, the momentum and kinetic energy of the system will be conserved. Since no external forces act on the helmet, the horizontal velocity component of the helmet (in the direction of the original motion of B) will remain constant, while the vertical velocity may change due to gravitational acceleration. Therefore, the velocity of Skater B will remain unaffected.
Remember, these explanations describe the general steps to solve the problem. It is important to plug in the actual values given in the problem to obtain accurate answers.