02.07.2022 - 17:59

A block and a gun are firmly affixed to opposite ends of a long glider mounted on a frictionless air track. The block and gun are a distance L apart. The system is initially at rest. The gun is fired and the bullet leaves the muzzle with a velocity vb and

Question:

A block and a gun are firmly affixed to opposite ends of a long glider mounted on a frictionless air track. The block and gun are a distance L apart. The system is initially at rest. The gun is fired and the bullet leaves the muzzle with a velocity vb and impacts on the block, becoming embedded in it. The mass of the bullet is mb and the mass of the gun-glider-block system is mp.

a. What is the velocity of the glider immediately after the bullet is fired?
b. What is the velocity of the glider immediately after the bullet comes to rest in the block?
c. How far does the glider move while the bullet is in transit between the gun and the block?

Answers (0)
  • Neoma
    April 10, 2023 в 18:05
    a. Immediately after the bullet is fired, the glider has a velocity of -vb*mb/mp, where vb*mb is the momentum of the bullet and -vb*mb/mp is the momentum gained by the gun-glider-block system in the opposite direction to conserve momentum. b. Immediately after the bullet comes to rest in the block, the combined mass of the block and bullet is mp+mb and the velocity of the combined mass is v=vb*mb/(mp+mb) in the direction of the bullet's initial velocity. c. The distance the glider moves while the bullet is in transit can be calculated using the equation d=v*t, where v is the velocity of the glider and t is the time it takes for the bullet to travel from the gun to the block. This time can be calculated using the equation L=(vb*t)-(1/2)(vb*mb)/(mp+mb) * t^2, where the first term represents the distance the bullet travels and the second term represents the distance the combined mass of the block and bullet moves due to the recoil. Solving for t and substituting into the equation for d gives d=L-(vb*mb)/(mp+mb)*L/vb.
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