Balllistic Spinner A bullet of mass = 0.01 kg is moving with initial velocity Vb

Question

Balllistic Spinner A bullet of mass = 0.01 kg is moving with initial velocity Vb = 160 m/s and strikes a pendulum of mass M = 0.3 kg hanging from a massless string of length R = 50 cm. The bullet sticks in the pendulum so this is an inlastic collision, and the pendulum swings around the circle after the collision. What is the velocity v1 of the pendulum/bullet just after the bullet lodges in the pendulum, in m/s? Use conservation of energy to find the velocities V2 and v% of the pendulum in the diagram, in m/s. W hat is the maximum tension T on the pendulum string? W hat are the magnitude and direction of the total acceleration acting on the mass at the location marked by v2?

Explanation / Answer

Apply Law of Conservation of Momentum

Inital Momentum = Final Momentum

Therefore

0.01*160 = (0.3 + 0.01)*V

V = 5.16 m/sec

0.5*m*V^2 = mgR + 0.5*m*v2^2

0.5*5.16^2 = 9.8*0.50 + 0.5*v2^2

v2 = 4.10 m/sec

Also

0.5*m*V^2 = mg2R + 0.5*m*v3^2

0.5*5.16^2 = 9.8*0.50*2 + 0.5*v3^2

v3 = 2.65 m/sec

Maximum Tension , T = Mg + +MV^2/r

= (0.3+0.01)*9.8 + (0.3+0.01)*5.16^2/0.5

= 19.5458 N

Net Acceleration = sqrt((v2^2/r)^2 + g^2)

= sqrt((4.10^2/0.5)^2 + 9.8^2)

= 35.019 m/sec^2

Let it makes an angle with the vertical

= arc tan((4.10^2/0.5)/9.8)

= 73.748 degree

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