A block of mass = 2.20 kg slides down a 30.0 deg incline which is 3.60 m high. At the bottom, it strikes a block of mass = 7.00 kg which is at rest on a horizontal surface . (Assume a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored.
Then I determined the speeds of both blocks after the collision:
m1=4.38
m2=4.02
Now I have to....
Determine how far back up the incline the smaller mass will go.
How did you figure out m2?
To determine how far back up the incline the smaller mass will go, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of a system remains constant as long as there are no external forces acting on it.
Step 1: Determine the initial mechanical energy of the system before the collision.
The initial mechanical energy of the system includes the potential energy due to the height of the block on the incline and the kinetic energy of the block sliding down. Since the block is at rest before sliding down, its initial kinetic energy is zero.
Initial potential energy = mass * gravitational acceleration * height
Initial potential energy of the smaller mass (m1) = 2.20 kg * 9.8 m/s^2 * 3.60 m
Step 2: Determine the final mechanical energy of the system after the collision.
Since the collision is elastic, both blocks will rebound after the collision with the same speed. We can use this speed to determine the total kinetic energy of the system after the collision.
Total kinetic energy after collision = (1/2) * (m1 + m2) * velocity^2
Step 3: Determine the height at which the smaller mass will stop.
Final potential energy = mass * gravitational acceleration * height
Final potential energy of the smaller mass (m1) = 2.20 kg * 9.8 m/s^2 * height
Step 4: Equate the initial mechanical energy to the final mechanical energy and solve for the height.
Initial potential energy of m1 = Final potential energy of m1 + Final potential energy of m2 + Total kinetic energy after collision
2.20 kg * 9.8 m/s^2 * 3.60 m = 2.20 kg * 9.8 m/s^2 * height + 7.00 kg * 9.8 m/s^2 * height + (1/2) * (2.20 kg + 7.00 kg) * velocity^2
Now, substitute the given values for the masses and velocity:
2.20 kg * 9.8 m/s^2 * 3.60 m = 2.20 kg * 9.8 m/s^2 * height + 7.00 kg * 9.8 m/s^2 * height + (1/2) * (2.20 kg + 7.00 kg) * (4.02 m/s)^2
Solving this equation will give you the height at which the smaller mass will stop back up the incline.