in a closed system, an object with the mass of 10kg moves at a velocity of 5.4 m/s. it collides with a second object that has a mass of 12 kg and that is moving in the opposite direction with a velocity of 3.5 m/s. what is the total momentum of the system after the collision.
To find the total momentum of the system after the collision, we need to use the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and velocity. It is given by the formula:
Momentum = mass × velocity
Let's calculate the momentum of the first object:
Momentum of the first object = mass × velocity
= 10 kg × 5.4 m/s
= 54 kg·m/s
Now, let's calculate the momentum of the second object:
Momentum of the second object = mass × velocity
= 12 kg × (-3.5 m/s) (Note that the second object is moving in the opposite direction, so we use a negative sign for velocity)
= -42 kg·m/s
To find the total momentum of the system after the collision, we simply add the momenta of the two objects:
Total momentum = Momentum of the first object + Momentum of the second object
= 54 kg·m/s + (-42 kg·m/s)
= 12 kg·m/s
Therefore, the total momentum of the system after the collision is 12 kg·m/s.
one object is traveling northeast with a momentum of 12kg-m/s, while second object is traveling southwest on the same path as the first object with a momentum of 4kg-m/s. when the two objects collide, they stick together. what is the momentum after their collision
To find the momentum after the collision, we need to consider both the magnitudes and the directions of the momenta.
Let's start by breaking down the momenta into their vector components using the directions of northeast and southwest.
The momentum of the first object traveling northeast can be broken down into its x and y components:
Momentum1x = 12 kg-m/s * cos(45°) ≈ 8.49 kg-m/s (in the positive x direction)
Momentum1y = 12 kg-m/s * sin(45°) ≈ 8.49 kg-m/s (in the positive y direction)
The momentum of the second object traveling southwest can also be broken down into its x and y components:
Momentum2x = 4 kg-m/s * cos(225°) ≈ -2.83 kg-m/s (in the negative x direction)
Momentum2y = 4 kg-m/s * sin(225°) ≈ -2.83 kg-m/s (in the negative y direction)
When the two objects collide and stick together, the x and y components of momentum add up to give the total momentum of the system after the collision.
Total momentumx = Momentum1x + Momentum2x
≈ 8.49 kg-m/s + (-2.83 kg-m/s)
≈ 5.66 kg-m/s (in the positive x direction)
Total momentumy = Momentum1y + Momentum2y
≈ 8.49 kg-m/s + (-2.83 kg-m/s)
≈ 5.66 kg-m/s (in the positive y direction)
To find the magnitude and direction of the total momentum, we can use the Pythagorean theorem and trigonometry:
Total momentum = sqrt((Total momentumx)^2 + (Total momentumy)^2)
≈ sqrt((5.66 kg-m/s)^2 + (5.66 kg-m/s)^2)
≈ sqrt(32.08 kg^2-m^2/s^2 + 32.08 kg^2-m^2/s^2)
≈ sqrt(64.16 kg^2-m^2/s^2)
≈ 8 kg-m/s
The direction of the total momentum can be found using the inverse tangent:
Direction = atan(Total momentumy / Total momentumx)
≈ atan(5.66 kg-m/s / 5.66 kg-m/s)
≈ atan(1)
≈ 45°
Therefore, the momentum after the collision is approximately 8 kg-m/s to the northeast.
a football kick returner catches the ball just as the player from the opposing team dives to tackle him. at the time of impact the returners momentum is 0kg-m/s, and the diving players momentum is 130kg-m/s. what will the magnitude of the total momentum be just after the collision
To find the magnitude of the total momentum just after the collision, we need to consider the conservation of momentum principle.
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
Given that the momentum of the returner is 0 kg-m/s and the momentum of the diving player is 130 kg-m/s, the total momentum before the collision is:
Total momentum before collision = Momentum of returner + Momentum of diving player
= 0 kg-m/s + 130 kg-m/s
= 130 kg-m/s
Now, since the total momentum before the collision is equal to the total momentum after the collision, the magnitude of the total momentum after the collision will also be 130 kg-m/s.
howard leans on the wall with a force of 50 n of force. according to newtons third law of motion what force is being applied and in what direction
According to Newton's third law of motion, for every action, there is an equal and opposite reaction.
In this case, Howard is leaning on the wall with a force of 50 N. According to Newton's third law, the wall will exert a force back on Howard that is equal in magnitude but opposite in direction to the force Howard is applying.
Therefore, the force being applied by the wall on Howard is also 50 N, but in the opposite direction to the force Howard is applying.