The ballistic pendulum apparatus consists of a launcher, a projectile, and a pen

Question

The ballistic pendulum apparatus consists of a launcher, a projectile, and a pendulum. Attached to the swinging end of the pendulum is a bucket designed to catch the projectile and hold it, creating an inelastic collision. The bucket rests directly in front of the launcher, right at the exit point of the projectile (fig. 1). figure 1 Once the launch mechanism is released, the projectile enters the bucket and the arm swings up, moving past a grooved plastic strip. A ratchet mechanism on the bucket slides past each groove, allowing motion up but not back down. Once the pendulum stops (kinetic energy is fully transferred to potential), it stays locked in place recording the maximum height of the swing. Our measurements consist of the initial height of the center of mass (CoM) of the pendulum, and a series of heights taken from several trial launches. This data is presented for analysis in the next section.

Blackwood Ballistic Pendulum Experiment

1. Is the kinetic energy conserved? If not, what evidence from your calculations (calculations have no been done since lab will not be performed until next week in class) do you have? Explain

2. Is momentum conserved? Explain

3. What is the velocity of the pendulum before the ball strikes it?

Explanation / Answer

1 ans

the kinetic energy is not conserved because this experiment is inelastic collision the kinetic energy of before collision is not equal to kinetic energy of after collision so the kinetic energy is not exist

(2) ans

the momentum is conserved

because m1u1+m2u2=[m1+m2]V

m1u1+0=[m1+m2]v

common velocity V=m1u1/m1+m2

so the momentum of before collision is equal to momentum of after collision

3 ans

k.e of the system after collision is given by

K.E=1/2[m1+m2]V^2

P.E at highest point =[m1+m2]gh

K.E at lowest point =P.E at highest point

1/2 [m1+m2]v^2=[m1+m2]gh

V=[2gh]^1/2

so the velocity of the pendulum u1={[m1+m2]/m1}*(2gh)^1/2

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