Answer to Question #105120 in Classical Mechanics for Jerico

Question

1. Let's combine momentum and collisions with some of the other things we've learned. A small block of mass M = 0.201 hangs on the end of a light, in extensible string, length R = 25 cm. A small dart (of mass m = 0.10 kg and small enough to be considered a particle) traveling in the horizontal direction collides with and remains fixed in the block. What is the minimum speed v of the dart such that the combined object completes a circular path around the support point of the block? (Hint: Consider the different parts of this question and analyze each separately before combining them. How fast must the block be traveling at the top? And don't forget significant figures.)

Minimum speed = _____

Answer 1

Assume that before the interaction the potential energy of the block is 0. After the dart sticks to the block, the block also acquires the kinetic energy and they move as one, meanwhile the kinetic energy converts to the potential energy and some other kinetic energy necessary to move along the circle:

$$KE=\frac{1}{2}(m+M)u^2=PE+KE_2.$$

According to the condition, the block and the dart must go up for a height of 2R, that is, acquire the potential energy of

$$PE=2(m+M)gR.$$

According to law of conservation of energy (just substitute everything):

$$\\\frac{1}{2}(m+M)u^2=2(m+M)gR+\frac{1}{2}(m+M)v_r^2,$$

where $v_r$ - speed required to "neutralize" the acceleration due to gravity and provide the circular motion:

$$(m+M)\frac{v_r^2}{R}=(M+m)g,\\ \space\\ v_r^2=gR.$$

Therefore:

$$\\\frac{1}{2}(m+M)u^2=2(m+M)gR+\frac{1}{2}(m+M)gR,\\ \space\\ u=\sqrt{5gR}.$$

How do we find the speed of the dart v? Remember about momentum conservation:

$$mv=(m+M)u,\\ \space\\ v=\frac{m+M}{m}u.$$

Substitute:

$$u=\frac{m+M}{m}\sqrt{5gR}=11\text{ m/s}.$$