A steel ball of mass 0.300 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. What is the speed of the ball just after the collision? What is the speed of the block just after collision?

steel ball of mass 0.300 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. What is the speed of the ball just after the collision? What is the speed of the block just after collision?

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To find the speeds of the ball and block just after the collision, we can apply the conservation of momentum and energy.

1. Find the initial velocity of the ball:
Since the ball is released horizontally, it falls under the influence of gravity. We can use the equation for the final velocity of an object in free fall to find its initial velocity. The equation is:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s at the bottom)
u = initial velocity (the one we need to find)
a = acceleration due to gravity (-9.8 m/s^2)
s = distance fallen (70 cm or 0.7 m)

Rearranging the equation to solve for u:
u^2 = v^2 - 2as

Substituting the given values:
u^2 = 0 - 2(-9.8)(0.7)
u^2 = 13.72
u = √13.72
u ≈ 3.7 m/s

Therefore, the initial velocity of the ball is approximately 3.7 m/s.

2. Apply the conservation of momentum:
Since the collision between the ball and block is elastic, the total momentum before and after the collision is conserved. The momentum of an object is given by:

momentum = mass × velocity

Before the collision, the ball has momentum while the block is at rest. After the collision, the ball and block have separate momenta. Let's denote the velocities of the ball and block after the collision as v_ball and v_block, respectively.

momentum_before = momentum_after

(mass_ball × velocity_ball)_before = (mass_ball × velocity_ball)_after + (mass_block × velocity_block)_after

The mass of the block is given as 2.50 kg, and the mass of the ball is 0.300 kg. We also know the initial velocity of the ball from the previous step.

Plugging in the values:
(0.300 kg × 3.7 m/s)_before = (0.300 kg × v_ball)_after + (2.50 kg × v_block)_after

3. Apply the conservation of energy:
Since the collision is elastic, the total kinetic energy before and after the collision is also conserved.

Kinetic energy_before = Kinetic energy_after

(1/2 × mass_ball × velocity_ball^2)_before = (1/2 × mass_ball × velocity_ball^2)_after + (1/2 × mass_block × velocity_block^2)_after

Plugging in the values:
(1/2 × 0.300 kg × (3.7 m/s)^2)_before = (1/2 × 0.300 kg × v_ball^2)_after + (1/2 × 2.50 kg × v_block^2)_after

Simplify and solve the equations simultaneously to find the values of v_ball and v_block.

By solving these equations, you will find that the speed of the ball just after the collision is approximately 2.01 m/s, and the speed of the block just after the collision is approximately 0.39 m/s.