1. A bullet of mass 6 g strikes a ballistic pendulum of mass 2.0 kg. The center

Question

1. A bullet of mass 6 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance of 12 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed. m/s

2. In the figure below, block 1 of mass m1 slides along an x axis on a frictionless floor with a speed of 1.60 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 2.30m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 2.30m2.

(a) Express the speed of block 3 just after the collision in terms of the initial speed of block 1, v1i. Note whether it is greater than, less than, or equal to the initial speed of block 1.
v3 =    v1i

(b) Express the kinetic energy of block 3 just after the collision in terms of the initial kinetic energy of block 1, K1i. Note whether it is greater than, less than, or equal to the initial kinetic energy of block 1.
K3 =   K1i

(c) Repeat for the magnitude of momentum of block 3 just after the collision, noting whether it is greater than, less than, or equal to the initial magnitude of momentum of block 1.
p3 =   p1i

Explanation / Answer

1)
Take v as the velocity of the bullet and m as the mass
Initial momentum of the bullet = mv
Consider that the bullet + pendulum moves with an initial velocity V and the mass of the pendulum as M
Momentum just after the collision = (m + M)V
Equating initial and the final momentum,
mv = (m+M)V
v = [(m+M)V]/m ..(1)

The bullet + pendulum rises to a height of 12 cm = 0.12 m.
Using the conservation of energy,
1/2(m+M)V2 = (m+M)gh
V = sqrt[2gh]
= sqrt[2 x 9.8 x 0.12]
= 1.534 m/s

Substituting in (1)
v = [2.006 x 1.534] / 0.006
= 512.74 m/s
________________________________________________

2)
Consider the first two masses.
Initial momentum of system = m1v1
Take v1' as the velocity of mass m1 after collision and v2 is the velocity of the mass m2 after collision.
Final momentum of the system = -m1v1' + m2v2
Equating both momentum,
m1v1 = - m1v1' + 2.3m1v2
v1 = - v1' + 2.3v2
v1' = 2.3v2 - v1

Initial kinetic energy = 1/2 m1v12
Final kinetic energy = 1/2 m1(v1')2 + 1/2m2(v2)2
Since the collision is elastic, initial KE = final KE
1/2 m1v12 = 1/2 m1(v1')2 + 1/2m2(v2)2
v12 = (v1')2 + 2.3(v2)2
Substituting for v1'
v12 = (2.3v2 - v1)2 + 2.3(v2)2.
2.3v22 - 2v1v2 + v22 = 0
3.3v2 = 2v1
v2 = 2/3.3v1
= (2/3.3) x 1.6
= 0.9697 m/s
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Consider the second and third two masses.
Initial momentum of system = m2v2
Take v2' as the velocity of mass m2 after collision and v3 is the velocity of the mass m3 after collision.
Final momentum of the system = -m2v2' + m3v3
Equating both momentum,
m2v2 = - m2v2' + 2.3m2v3
v2 = - v2' + 2.3v3
v2' = 2.3v3 - v2

Initial kinetic energy = 1/2 m2v22
Final kinetic energy = 1/2 m2(v2')2 + 1/2m3(v3)2
Since the collision is elastic, initial KE = final KE
1/2 m2v22 = 1/2 m2(v2')2 + 1/2m3(v3)2
v22 = (v2')2 + 2.3(v3)2
Substituting for v1'
v22 = (2.3v3 - v2)2 + 2.3(v3)2.
2.3v32 - 2v2v3 + v32 = 0
3.3v3 = 2v2
v3 = 2/3.3v2
= (2/3.3) x 0.9697
= 0.5877 m/s
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a)
v3 = 0.5877 m/s
= n times 1.6 m/s
= 0.3673 v1

b)
K3 = 1/2 m3(v3)2
= 0.5 x 2.32m1 x 0.58772
= 0.5m1 x 1.827
K1 = 0.5 x m1 x 1.62
= 0.5m1 x 2.56
K3 = 0.714 K1

c)
P3 = m3v3
= 2.32m1 x0.5877
= 3.11
P1 = m1v1
= 1.6m1
P3 = 1.943 P1

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