Question

4. A steel ball of mass m is fastened to a light cord of length L and released when the cord is horizontal. At the
bottom of its path, the ball strikes a hard plastic block of mass M = 3m, initially at rest on a frictionless
surface. The collision is elastic.
(a) Find the tension in the cord when the ball’s height above its lowest position is L/3. Write your answer
in terms of m and g.
(b) Find the speed of the block immediately after the collision.
(c) To what height h will the ball rebound after the collision?

Answer 1

Answer on Question 49101, Physics, Mechanics | Kinematics | Dynamics

Question:

4. A steel ball of mass $m$ is fastened to a light cord of length $L$ and released when the cord is horizontal. At the bottom of its path, the ball strikes a hard plastic block of mass $M = 3m$ , initially at rest on a frictionless surface. The collision is elastic.

(a) Find the tension in the cord when the ball's height above its lowest position is L/3. Write your answer in terms of m and g.

(b) Find the speed of the block immediately after the collision.

(c) To what height h will the ball rebound after the collision?

Solution:

a) Let's draw a free-body diagram:

So, let's write the forces acting on the ball:

$$F _ {T} - m g \sin \alpha = F _ {\text {c e n t r i p e t a l}} = \frac {m v ^ {2}}{L}.$$

We can see that we have two unknowns in this equation $m v^2$ and $\sin \alpha$ .

Let's write the law of conservation of energy to find $m v^2$ . Then we have:

$$K E _ {i} + P E _ {i} = K E _ {f} + P E _ {f},$$ $$0 + m g L = \frac {1}{2} m v ^ {2} + m g \frac {L}{3},$$ $$m v ^ {2} = \frac {4}{3} m g L.$$

From the triangle in the free-body diagram we can see that $\sin \alpha = \frac{2}{3} L / L = \frac{2}{3}$ .

So, after substituting $mv^2$ and $\sin \alpha$ into first equation we obtain:

$$F _ {T} = \frac {2}{3} m g + \frac {4}{3} m g = 2 m g.$$

b) First we find the velocity of the ball before it hits the block. Let's write the law of conservation of energy:

$$K E _ {i} + P E _ {i} = K E _ {f} + P E _ {f},$$ $$0 + m g L = \frac {1}{2} m v ^ {2} + 0,$$ $$v = \sqrt {2 g L}.$$

So, because we have elastic head-on collision and kinetic energy is conserved we can obtain the velocity of the block after collision:

$$v _ {2} ^ {\prime} = \frac {2 m _ {1}}{m _ {1} + M} v = \frac {2 m _ {1}}{m _ {1} + 3 m _ {1}} \sqrt {2 g L} = \frac {2 m _ {1}}{4 m _ {1}} \sqrt {2 g L} = \frac {1}{2} \sqrt {2 g L}.$$

c) We can find $h$ from the law of conservation of energy:

$$K E _ {i} + P E _ {i} = K E _ {f} + P E _ {f},$$ $$\frac {1}{2} m v _ {1} ^ {\prime 2} + 0 = 0 + m g h,$$ $$h = \frac {v _ {1} ^ {\prime 2}}{2 g}.$$

We can find the velocity of the ball after collision from the formula:

$$v _ {1} ^ {\prime} = \frac {m _ {1} - M}{m _ {1} + M} v _ {1} = \frac {m _ {1} - 3 m _ {1}}{m _ {1} + 3 m _ {1}} \sqrt {2 g L} = - \frac {2 m _ {1}}{4 m _ {1}} \sqrt {2 g L} = - \frac {1}{2} \sqrt {2 g L}.$$

After substituting $v_{1}^{\prime}$ into the formula for $h$ we obtain:

$$h = \frac {\left(- \frac {1}{2} \sqrt {2 g L}\right) ^ {2}}{2 g} = \frac {\frac {1}{4} 2 g L}{2 g} = \frac {1}{4} L.$$

Answer:

a) $F_{T} = 2mg$

b) $v_{2}^{\prime} = \frac{1}{2}\sqrt{2gL}$

c) $h = \frac{1}{4} L$

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