In the figure below, a 3.7 kg box of running shoes slides on a horizontal frictionless table and collides with a 2.0 kg box of ballet slippers initially at rest on the edge of the table, at height h = 6.9 m. The speed of the 3.7 kg box is 5.0 m/s just before the collision. If the two boxes stick together because of packing tape on their sides, what is their speed just before they strike the floor?
initial momentum = 3.7 * 5 = 18.5 kg m/s
mass after crash = 3.7 + 2.0 = 5.7 kg
so
18.5 = 5.7 u
u = 3.25 m/s which is the horizontal speed until it hits the floor.
NOW I DO NOT BELIEVE that your table is 6.9 meters high so I will just outline the rest
How long to fall h meters?
h = 4.9 t^2
solve for t, time in air
what is vertical speed at floor?
v = - g t = -9.8 t
so speed at floor = sqrt (v^2 + 3.25^2)
To solve this problem, we need to apply the principle of conservation of mechanical energy. Conservation of mechanical energy states that the total mechanical energy of a system remains constant if no external forces do work on the system.
Before the collision, the only form of mechanical energy in the system is the kinetic energy of the 3.7 kg box, which is given by:
KE1 = (1/2) * m1 * v1^2
Where m1 is the mass of the 3.7 kg box and v1 is its speed just before the collision.
After the collision, the two boxes stick together and fall to the floor. When they reach the floor, all of their initial kinetic energy will be converted into potential energy due to their height above the ground.
The potential energy of the system when it reaches the floor is given by:
PE = m1 * g * h
Where m1 is the mass of the combined boxes and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the total mechanical energy is conserved, we can equate the kinetic energy before the collision (KE1) to the potential energy after the collision (PE).
(1/2) * m1 * v1^2 = m1 * g * h
Now, let's solve for the speed just before the boxes strike the floor.
First, rearrange the equation:
v1^2 = 2 * g * h
Substitute the given values:
v1^2 = 2 * 9.8 * 6.9
v1^2 = 135.48
Finally, take the square root of both sides to find the speed:
v1 ≈ √135.48
v1 ≈ 11.63 m/s
Therefore, the speed of the combined boxes just before they strike the floor is approximately 11.63 m/s.