Block-Bullet System

Related BrainMass Solutions

• Energy Conservation in Oscillatory Motion

  We can get So, the initial speed of the bullet is 6.426(m/s) Now, look at the bullet-block spring system, the mass is m+M=0.51kg.

• Bullet Motion and Conservation of Energy

  In our case, right after the bullet hits the block, the initial potential energy of the system is zero. When the block-bullet system reaches the maximal height the final kinetic energy is zero (the velocity is zero).

• Inelastic collision and ballistic pendulum

  When a bullet ( of mass 'm') hits a Block ( of mass M) with a speed 'v' and get arrested in the block, the bullet and block system will start moving with a speed 'Vb'.

• Speed of a bullet measured using ballistic pendulum !

  The momentum of the bullet (m1v1) before the collision with the block has to equal the momentum of the block-bullet system after the collision. The collision between the block and the bullet is inelastic, so kinetic energy (KE) is not conserved.

• collision between a bullet and a wooden block of a pendulum

  102501 collision between a bullet and a wooden block of a pendulum A 2.50 g bullet, traveling at a speed of 425 m/s, strikes the wooden block of a ballistic pendulum. The block has a mass of 215 g.

• bullet hitting a wooden block

  for a bullet hitting a wooden block.

• Elastic and Inelastic Collisions

  Let v be the speed of the bullet and V be the velocity of the block just after the bullet gets in to it.

• Mechanics: 3 Probles on conservation of energy and momentum

  Part C What is the speed of the block/bullet system after the collision?

• Physics: Momentum and Speed

  a) Initial momentum of the bullet = 0.005 x 109.5 = 0.548 kg.m/s Initial momentum of the block = 0 As the bullet gets embedded in the block, the velocity of both will be the same after the collision. Let the common speed be v.