A system containing a frictionless pulley is described in the diagram on the right (the 2.0 kg box is on the floor while the 3.0 kg box is 0.5m high) . If this system is released, what will be the momentum of:
A. the 2.0 kg box when the 3.0 kg box hits the floor?
b. the 3.0 kg when it hits the floor?
a. I would work this by energy.
mass2 gains PE, and gains KE, the mass3 loses PE, and gainsKE. the net change in total energy is zero.
Both masses will have the same velocity.
1/2 2 V^2+2g*.5-3g*.5+1/2 3 v^2=0
v^2(1+3/2)=-g+1.5g
v= sqrt (.5g*2/3)=sqrt(g/3)
check that.
To find the momentum of the boxes in this system, we need to consider the conservation of momentum. The total momentum of the system before and after the event should be the same, assuming there are no external forces acting on the system.
Now let's find the momentum of each box:
A. The 2.0 kg box: When the 3.0 kg box hits the floor, the 2.0 kg box will be at rest. Initially, it has zero velocity, so the momentum is zero.
B. The 3.0 kg box: To determine the momentum just before it hits the floor, we need to find its velocity. Recall that the potential energy lost by the 3.0 kg box as it falls is converted into kinetic energy. We can use the conservation of energy to find the velocity.
The potential energy (PE) of the 3.0 kg box at a height of 0.5m is given by:
PE = m * g * h
where m is the mass (3.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (0.5 m). Substituting the values, we have:
PE = 3.0 kg * 9.8 m/s^2 * 0.5 m = 14.7 J
This potential energy is converted into kinetic energy just before the box hits the floor. The kinetic energy (KE) is given by:
KE = 0.5 * m * v^2
where v is the velocity of the box just before impact. Substituting the known values, we have:
14.7 J = 0.5 * 3.0 kg * v^2
v^2 = (2 * 14.7 J) / 3.0 kg
v^2 = 29.4 J / 3.0 kg
v^2 = 9.8 m^2/s^2
Taking the square root of both sides, we get:
v = √9.8 m^2/s^2
v = 3.1 m/s
Now that we have the velocity, we can calculate the momentum of the 3.0 kg box just before hitting the floor. Momentum (p) is given by:
p = m * v
p = 3.0 kg * 3.1 m/s = 9.3 kg m/s
Therefore, the momentum of the 3.0 kg box just before hitting the floor is 9.3 kg m/s.