The bird perched on the swing has a mass of 48.0 g, and the base of the swing has a mass of 159 g. The swing and bird are originally at rest, and then the bird takes off horizontally at 1.20 m/s. How high will the base of the swing rise above its original level? Disregard friction.
first use conservation of momentum to get swing speed at bottom
Vswing = Vbird (Mass bird/Mass swing)
then
m g h = (1/2) m Vswing^2
or
h = Vswing^2/(2 *9.81)
To solve this problem, we need to apply the principle of conservation of momentum.
First, let's find the total initial momentum of the system, which includes both the bird and swing base. Since the swing and bird are originally at rest, their initial velocities are zero. Thus, the initial momentum is given by:
Initial Momentum = mass * initial velocity
For the bird:
Bird's initial momentum = (mass of bird) * (initial velocity of bird)
= 48.0 g * 0 m/s (since the bird is at rest)
= 0 kg·m/s
For the swing base:
Swing base's initial momentum = (mass of swing base) * (initial velocity of swing base)
= 159 g * 0 m/s (since the swing base is at rest)
= 0 kg·m/s
Since both the bird and the swing base are initially at rest, the total initial momentum of the system is also zero.
Next, let's consider the system after the bird takes off horizontally. The bird will gain some horizontal momentum, but there will be an equal and opposite horizontal momentum gained by the swing base. As a result, the total momentum of the system will remain zero, in accordance with the principle of conservation of momentum.
Now, let's find the height the base of the swing will rise above its original level. Since the total momentum of the system remains zero, the vertical motion of the swing base will not be affected by the bird's takeoff.
Therefore, the height the base of the swing will rise above its original level is zero.