A bullet of mass 0.1kg initially travels with a velocity of 200m/s in the positi

Question

A bullet of mass 0.1kg initially travels with a velocity of 200m/s in the positive x direction. It strikes and embeds itself in a block of mass 10kg which was initially at rest and hanging from a light rope.

1) Immediately after the bullet strikes, what is the tension in the rope?

2) When the block has swung through an angle of 20 degrees, how much work has gravity done on it?

3) When the block has swung through an angle of 20 degrees, what is the speed of the block?

4) When the block has swung through an angle theta as illustrated (theta is between the positive x-axis and negative y-axis), what are the directions of the angular momentum and torque, both measured about the location where the rope is attached to the ceiling?

Thanks

Explanation / Answer

from momentum coservation

initial momentum = final momentum

m*u = (m+M)*V

V = (m*u)/(M+m) = (0.1*200)/(0.1+10) = 1.98 m/s

T - (m+M) g = (M+m)*v^2/L

T = (10.1*1.98*1.98)/1 + (10.1*9.8)

T = 138.57604 N

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2) Wg = m*g*L*(1-cos20)

Wg = 10.1*9.8*1*(1-cos20)

Wg = 5.969 J

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3) 0.5*(m+M)*V^2 = wg + 0.5*m*vf^2

Vf = 1.65 m/s

4) L = m*v*L = 10.1*1.65*1 = 16.665 kg m/s

torque = (M+m)*g*l*sin20 = 33.85 Nm

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