steel ball of mass 0.310 kg is fastened to a cord that is 86.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 3.10 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
To find the speed of the ball and the block after the collision, we need to use the principles of conservation of energy and conservation of momentum.
(a) Calculating the speed of the ball:
1. We can start by calculating the gravitational potential energy of the ball at its initial position by using the formula:
Potential Energy = m * g * h
where m is the mass of the ball (0.310 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the ball (which is equal to the length of the cord, 86.0 cm = 0.86 m).
Potential Energy = (0.310 kg) * (9.8 m/s²) * (0.86 m) = 2.6792 J
2. At the bottom of the path, the potential energy is converted into kinetic energy since the ball is moving horizontally. So, we can equate the potential energy to the kinetic energy:
Kinetic Energy = 1/2 * m * v²
where v is the speed of the ball.
2.6792 J = 1/2 * (0.310 kg) * v²
5.3584 J = (0.310 kg) * v²
Dividing both sides by (0.310 kg):
17.316 J/kg = v²
Taking the square root of both sides:
v = sqrt(17.316 J/kg) ≈ 4.165 m/s
So, the speed of the ball just after the collision is approximately 4.165 m/s.
(b) Calculating the speed of the block:
Since the collision is elastic, we can use the principle of conservation of momentum to find the speed of the block.
1. Initially, the block is at rest, so its initial momentum is zero (since momentum = mass * velocity).
2. The total momentum after the collision is equal to the momentum of the ball and the block.
Momentum of the ball before the collision = m_ball * v
Momentum of the block before the collision = m_block * 0 (initially at rest)
Momentum of the ball after the collision = m_ball * v_after
Momentum of the block after the collision = m_block * v_block_after (what we need to find)
3. Since momentum is conserved, we can set up the equation:
m_ball * v = m_ball * v_after + m_block * v_block_after
Substituting the known values:
(0.310 kg) * (4.165 m/s) = (0.310 kg) * v_after + (3.10 kg) * v_block_after
Dividing both sides by 0.310 kg:
(4.165 m/s) = v_after + (3.10 kg / 0.310 kg) * v_block_after
(4.165 m/s) = v_after + 10 * v_block_after
Simplifying:
v_after + 10 * v_block_after = 4.165 m/s
4. Since the collision is elastic, the relative velocities of the ball and the block after the collision will be the same as before the collision but in the opposite direction. So, the relative velocity is (-4.165 m/s).
5. Now we can set up the second equation by considering the collision between the ball and the block.
Using the equation for conservation of kinetic energy:
1/2 * m_ball * v^2 = 1/2 * m_ball * v_after^2 + 1/2 * m_block * v_block_after^2
Simplifying:
(0.310 kg) * (4.165 m/s)^2 = (0.310 kg) * v_after^2 + (3.10 kg) * v_block_after^2
0.53416325 J = 0.310 kg * v_after^2 + (3.10 kg) * v_block_after^2
6. Combining the two equations:
v_after + 10 * v_block_after = 4.165 m/s
0.53416325 J = 0.310 kg * v_after^2 + (3.10 kg) * v_block_after^2
You can solve these two equations to find the values of v_after and v_block_after using numerical methods or an algebraic solver.
Therefore, to find the speed of the ball and the block after the collision (a) and (b), we need to solve the system of equations formed by the principles of conservation of momentum and conservation of kinetic energy.