Newton's cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes one ball on the opposite end to be released. Which type of force can decrease the total momentum of this system ? (1 point) electrostatic force magnetic force friction force strong nuclear force
friction force
Are you sure?
Yes, I am sure. Friction force can decrease the total momentum of the system, as it opposes the motion of the metal balls and can cause them to slow down and eventually stop, reducing the total momentum of the system.
One object is traveling northeast with a momentum of 12 kg-m/s, while a second object is traveling southwest on the same path as the first object with a momentum of 4 kg-m/ s. When the two objects collide, they stick together . What is the momentum after their collision ? (1 point) 16 kg-m/ s southwest 8 kg-m/s northeast 8 kg-m/s southwest 16 kg-m/s northeast
The momentum of the system is conserved, so the total initial momentum of the two objects is:
12 kg-m/s (northeast) + 4 kg-m/s (southwest) = 8 kg-m/s (northeast)
After they collide and stick together, the total mass of the two objects is:
m = m1 + m2
Since the objects stick together, their velocities after the collision are the same and can be represented as V. Therefore, the total momentum after the collision is:
P = mV
The total momentum must be equal to the initial momentum:
8 kg-m/s (northeast) = (m1 + m2) V
We know that the total mass of the two objects is m1 + m2, so we can substitute that for m:
8 kg-m/s (northeast) = m V
Now we need to find the mass of the two objects. We don't have enough information to do this directly, but we can use the fact that the objects stick together to solve for the relative velocity before the collision.
We can represent the velocities of the two objects before the collision as v1 and v2. Since they are on the same path, but traveling in opposite directions, we know that their relative velocity before the collision is:
v_rel = |v1 - v2|
We aren't given the speeds of the objects, but we know that:
|v1| = |v2|
Since they are traveling in opposite directions on the same path, they must have equal speeds. Therefore, we can simplify the equation to:
v_rel = 2|v|
Now we can use the fact that the collision is perfectly inelastic (the objects stick together) to relate the relative velocity before the collision (v_rel) to the velocity after the collision (V):
V = v_rel/2
The relative velocity can be calculated from the momenta before the collision:
12 kg-m/s (northeast) + 4 kg-m/s (southwest) = m1 v1 + m2 v2
We don't know the individual masses, but since the objects stick together we can simplify the equation to:
16 kg-m/s northeast = (m1 + m2) v
Now we have two equations that relate the initial and final momenta:
8 kg-m/s northeast = (m1 + m2) V
16 kg-m/s northeast = (m1 + m2) v
We can solve for v and V in terms of m1 and m2:
v = 16 kg-m/s / (m1 + m2)
V = 8 kg-m/s / (m1 + m2)
Now we can set v equal to V and solve for m1 + m2:
16 kg-m/s / (m1 + m2) = 8 kg-m/s / (m1 + m2)
Multiplying both sides by (m1 + m2):
16 kg-m/s = 8 kg-m/s
This is a contradiction, so there is no solution. Therefore, our initial assumption that the two objects collide on the same path is incorrect. One possible explanation is that the two objects approach each other at an angle and collide off-center, causing them to move in a new direction after the collision. Without more information about the angle and location of the collision, we cannot determine the momentum after the collision.
Two identical merry -go -rounds are loaded up with equal amounts of mass . On merry -go -round A , the mass is placed in the center , while the mass is placed near the edges of merry -go -round B. The merry - go-rounds are pushed with equal forces , and merry -go-round A ends up moving with a greater angular velocity . Which statement helps explain this difference ? (1 point ) Merry -go -round A has a greater moment of inertia .
Merry -go -round A has more inertia .
Merry -go - round B has a greater moment of inertia .
Merry -go - round B has more inertia .
Merry-go-round A has a greater moment of inertia.
The moment of inertia of an object depends on its mass and the distance between the mass and the axis of rotation. In this case, the mass in merry-go-round A is placed in the center, which is closer to the axis of rotation than the mass in merry-go-round B, which is placed near the edges. This means that merry-go-round A has a smaller moment of inertia than merry-go-round B.
According to Newton's second law for rotational motion, the angular acceleration of an object is directly proportional to the torque applied to it and inversely proportional to its moment of inertia. Since the two merry-go-rounds are pushed with equal forces, the torque applied to them is the same. However, since merry-go-round A has a smaller moment of inertia, it will experience a greater angular acceleration than merry-go-round B. This means that it will reach a greater angular velocity in a shorter amount of time, explaining the difference in their final velocities.
Therefore, the statement "Merry-go-round A has a greater moment of inertia" helps explain the difference in their angular velocities.
In a closed system, an object with a mass of 10 kg moves at a velocity of 5.4m / 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 3.5m / s . What is the total momentum of the system after the collision ? (1 point) 42kg - m / s; 12kg - m / s O54 54kg - m / s; 66kg - m / s
Before the collision, the total momentum of the system is:
P1 = m1v1 + m2v2
where m1 = 10 kg, v1 = 5.4 m/s, m2 = 12 kg, and v2 = -3.5 m/s (negative because it is moving in the opposite direction). Plugging in the values, we get:
P1 = (10 kg)(5.4 m/s) + (12 kg)(-3.5 m/s)
P1 = 54 kg·m/s - 42 kg·m/s
P1 = 12 kg·m/s
The negative sign indicates that the momentum of the second object is in the opposite direction compared to the first object.
After the collision, the two objects stick together and move with a common velocity v. By conservation of momentum, the total momentum of the system remains the same:
P2 = (m1 + m2) v
where m1 + m2 = 22 kg (the total mass of the system). To find v, we need to use conservation of kinetic energy, since the collision is inelastic (some kinetic energy is lost in the collision). We can determine the kinetic energy of the system before and after the collision and equate them:
KE1 = (1/2) m1 v1^2 + (1/2) m2 v2^2
KE2 = (1/2)(m1 + m2) v^2
Plugging in the values, we get:
KE1 = (1/2)(10 kg)(5.4 m/s)^2 + (1/2)(12 kg)(-3.5 m/s)^2
KE1 = 81.54 J
KE2 = (1/2)(22 kg) v^2
We can assume that some kinetic energy is lost in the collision, so KE2 must be less than KE1. Let's say that KE2 is 80% of KE1 (to account for the lost energy):
KE2 = 0.8 KE1
KE2 = 0.8 (81.54 J)
KE2 = 65.24 J
Solving for v in the equation for KE2, we get:
v = sqrt((2 KE2) / (m1 + m2))
v = sqrt((2)(65.24 J) / (10 kg + 12 kg))
v = sqrt(4.27 m^2/s^2)
v = 2.07 m/s
Now we can calculate the total momentum of the system after the collision:
P2 = (10 kg + 12 kg)(2.07 m/s)
P2 = 44.52 kg·m/s
Therefore, the total momentum of the system after the collision is 44.52 kg·m/s.
Answer: 44.52 kg-m/s
In a closed system , an object with a mass of 10 kg 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 ? ( 1 point ) 42 kg - m / s
12 kg - m / s
54 kg - m / s
66 kg - m / s