3. m 30 Two metal spheres, suspended by vertical cords, initially just touch. Sp

Question

3. m 30 Two metal spheres, suspended by vertical cords, initially just touch. Sphere 1, with mass & is pulled to the left to a height h, 8.0 cm, and then released from rest. After swinging down, t undergoes an elastic collision with sphere 2, whose mass m, 75 g. What is the velocity viy f sphere 1 just after the collision? (Assume that the collision is elastic.) 4. A ballistic pendulum (see figure) is a device that was ppppy used to measure the speeds of "bullets" before electronic timing devices were available. It consists of a large block of wood of mass M, hanging from two long pairs of cords. A "bullet" of mass m is fired into the block and comes quickly to rest relative to the block. The block+ "bullet" combination swings upward, its center of mass rising a vertical distance h fore the pendulum comes momentarily to rest at the end of its arc. (a) What is the initial speed of the "bullet" if the block rises to a height of h? (b) What is the initial kinetic energy of the "bullet? How much of this energy remains as mechanical energy of the swinging pendulum? First solve using only variables, no numbers; then determine values ifM·5.4 kg, m . 9.5 g and h = 6.3 cm.

Explanation / Answer

3)
Initial velocity of the first metal sphere, vi = SQRT[2gh]
= SQRT[2 x 9.8 x 0.08] = 1.252 m/s

Consider vf1 be the velocity after collision for the first metal sphere and vf2 be the velocity of the second metal sphere after collision.
Consider m1 be the mass of first sphere and m2 be the mass of second sphere.
Initial momentum, P1 = m1 vi
Final momentum, P2 = m1 vf1 + m2 vf2
Using momentum conservation, P1 = P2
m1vi = m1 vf1 + m2 vf2
Substituting the values,
30 x 1.252 = 30 vf1 + 75 vf2
vf1 = 1.252 - 2.5 vf2      ...(1)

Using conservation of kinetic energy,
1/2 m1(vi)2 = 1/2 m1 (vf1)2 + 1/2 m2 (vf2)2.
Substituting for vi and dividing with m1,
1.568 = (vf1)2 + 2.5 (vf2)2
Substituting vf1 = 1.252 - 2.5 vf2
1.568 = [1.568 + 6.25 (vf2)2 - 6.26 vf2] + 2.5 (vf2)2
6.26 vf2 = 8.75 (vf2)2
vf2 = 6.26/8.75
= 0.716 m/s.

4)
Consider V be the initial velocity of the bullet + block system.
Using conservation of energy, 1/2 (M + m) V2 = (M + m) gh
V = SQRT[2gh]
= SQRT[2 x 9.8 x 0.063]
= 1.11 m/s.

a)
Consider v be the velocity of the bullet
Using conservation of momentum, mv = (m + M) V
v = [(m + M) V]/m
= [(5.4 + 0.0095) x 1.11] / 0.0095
= 632.75 m/s

b)
Kinetic energy of the bullet = 1/2 mv2
= 0.5 x 0.0095 x (632.75)2
= 1901.767 J

c)
Mechanical energy of the pendulum = 1/2 (M + m) V2 = (M + m) gh
= (5.4 + 0.0095) x 9.8 x 0.063
= 3.340 J
So only 3.34 J out of 1901.77 J remains as the mechanical energy of the pendulum.

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