A block of mass m = 2.60 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.80 kg which is at rest on a horizontal surface, Fig. 7-46. (Assume a smooth transition at the bottom of the incline, an elastic collision, and ignore friction.)
(a) Determine the speeds of the two blocks after the collision.
lighter block-
heavier block-
(b) Determine how far back up the incline the smaller mass will go.
To solve this problem, we can use the principles of conservation of energy and conservation of momentum.
(a) Determining the speeds of the two blocks after the collision:
1. Calculate the potential energy of the first block at the top of the incline.
Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
PE = 2.60 kg * 9.8 m/s^2 * 3.60 m
2. Calculate the kinetic energy of the first block at the bottom of the incline.
Since there is no friction, the total mechanical energy is conserved.
Kinetic energy (KE) = Potential energy at the top
KE = PE
3. Use the equation for kinetic energy:
KE = (1/2) * mass (m1) * velocity^2
where m1 is the mass of the first block.
4. Substitute the known values into the equation and solve for velocity:
(1/2) * 2.60 kg * velocity^2 = PE
velocity^2 = (2 * PE) / m1
velocity^2 = (2 * PE) / 2.60 kg
velocity = sqrt((2 * PE) / 2.60 kg)
5. Calculate the momentum of the first block using the equation:
Momentum (p1) = mass (m1) * velocity
6. Since the collision is elastic, the total momentum before the collision is equal to the total momentum after the collision.
7. Determine the momentum of the second block (p2) before the collision:
Momentum (p2) = mass (M) * velocity (v2)
Since the block is at rest, the initial velocity (v2) is zero.
8. Equate the initial momentum (p1) to the final momentum (p1' + p2) after the collision:
p1 = p1' + p2
p1' = p1 - p2
9. Determine the velocity of the second block (p1') after the collision:
p1' / m1 = velocity of the second block (p1' / m1)
p1' = (p1 - p2) / M
(b) Determining how far back up the incline the smaller mass will go:
1. Since the collision is elastic, the total mechanical energy is conserved.
The final kinetic energy of the first block should be equal to the potential energy it has when it reaches the maximum height on the incline.
2. Calculate the potential energy at the maximum height on the incline, using the same formula as in step 1 for part (a):
PE_max_height = m * g * h
PE_max_height = 2.60 kg * 9.8 m/s^2 * 3.60 m
3. Equate the final kinetic energy of the first block to its potential energy at the maximum height:
(1/2) * m * velocity^2 = PE_max_height
velocity^2 = (2 * PE_max_height) / m
velocity = sqrt((2 * PE_max_height) / m)
4. Assuming all the initial kinetic energy is converted to potential energy, the final velocity of the block at its maximum height is zero.
5. Use the equation for the conservation of mechanical energy:
Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy
6. Substitute the known values:
(1/2) * m * velocity^2 + PE = 0 + PE_max_height
7. Solve for the distance traveled up the incline (d):
d = (PE - PE_max_height) / m
By following these steps, you should be able to determine the speeds of the two blocks after the collision (part a) and how far back up the incline the smaller mass will go (part b). Just remember to substitute the known values into the equations and solve for the respective quantities.