A pendulum bob of mass \"3m\" is connected to a string of length \"L\". The bob

Question

A pendulum bob of mass "3m" is connected to a string of length "L". The bob is pulled back an angle of 90 degrees and given an initial velocity "v0" downward. When the pendulum bob reaches its lowers point it collides with the second pendulum bob. The second pendulum bob also of length "L" has a mass "m" and is initially at rest. The collision between the two bob's occur with perfect elasticity. After the collision the bob of mass "m" just makes it over the top without falling inward. Given [m,L]

a. The speed of the "m" mass when it swings over the top. (call it v4).

b. The speed of the "3m" mass just after the collision. (call it v3).

c. The speed of the "m" mass just after the collision. (call it v2).

d. The speed of the "3m" mass just before the collision. (call it v1).

e. The magnitude and direction of the acceleration of mass "3m" just before the collision.

f. The maximum angle the "3m" makes the vertical after the collision.

g. The initial velocity vo.

Explanation / Answer

(a)
At top,
m*g = m*v4^2/r
v4 = sqrt(g*L)

(b)
Speed of m just after collision,
Using Energy Conservation,
Initial Kinetic Energy = Final kinetic Energy + Final Potential Energy
1/2*m*v2^2 = 1/2 * m * v4^2 + m*g*(2*L)
1/2*v2^2 = 1/2*g*L + 2*g*L
v2 = sqrt(5*g*L)

Using Momentum Conservation,
Initial  Momentum = Final Momentum
3m*v1 = - 3m*v3 + m * v2
3*v1 + 3 * v3 = v2
v2 = 3*(v1+v3)

Initiak Kinetic Energy = Final Kinetic Energy
3m * v1^2 = 3m * v3^2 + m * v2^2
3*v1^2 = 3*v3^2 + v2^2
3*(v1 - v3) * (v1+v3) = v2^2
3*(v1 -v3) * v2/3 = v2^2
v1 - v3 = v2

v1 - v3 = 3*(v1+v3)
v1 - v3 = 3*v1 + 3*v3
- 2*v1 = 4*v3
v1 = - 2*v3

-2*v3 - v3 = v2
v3 = - v2/3
v3 = - sqrt(5*g*L)/3

(-ve sign mean the direction is to the left !!)

v1 = 2/3 *sqrt(5*g*L)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.