What Is The Speed Of The Blockbullet System After The Collision
Solved Figure 1 Of 1 What Is The Speed Of The Block Bullet Chegg Com
What is the speed of the blockbullet system after the collision. Energy to the block-bullet system between right after the collision and the systems swinging up 045m from the point of collision. A bullet of mass mb is fired horizontally with speed vi at a wooden block of mass mw resting on a frictionless table. Let be the initial velocity of the bullet.
The initial speed of the leading bumper car is 560 mathrmm mathrms and that of the trailing car is 600 mathrmm mathrms. So the speed of the system immediately after the collision is. Apply conservation of momentum to determine the relationship between v f and the bullets speed before the collision.
Consider the velocity before the collision to be v 0. Express your answer in terms ofi m and mb View Available Hint s. Momentum before the collision momentum after the collision mv o mMv f.
The speed of the bullet just before it hit the block was about 150 ms. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block the block with the bullet in it is traveling at speed vf.
14 Part a Enter an expression for the speed of the blockbullet system immediately after the collision in terms of defined quantities and g. Now this question can be solved in two ways-. Therefore the velocity of the center of mass of the system is 12 ms the velocity of the bullet block after the collision.
The bullet hits the block and becomes completely embedded within it. Putting it all together the correct way to account for the kinetic energy of the system would be. Maximum amount of kinetic energy is lose in these type of collision.
What is the speed of the blockbullet system after the collision. Also consider the velocity after the collision to be v f.
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If v 1 the velocity of the bullet and v 2 the initial velocity of the block 0 ms then.
After the collision the two-car system travels at speed at an angle east of north. What is the speed of the bullet block system after the collision. In completely still wind that 1422 ms bullet speed decays at about 10 meters per second each second so its peak is reached in less than 140010 seconds thats 140 seconds with NO air drag so you can guarantee that no bullet will reach 140 sec X 711 ms that is lower than 995 km high. We can rewrite the bullets speed relative to you in terms of the variables you care about. 10 pts Calculate the recoil velocity of a 60 kg rifle that shoots a 50. Lets say the velocity of the bullet after piercing the block be v1 ms the velocity of the block is v2 ms and mass does not change. What is the speed of the blockbullet system after the collision. Also consider the velocity after the collision to be v f. After the collision the blockbullet system swings and reaches a maximum height of h105 m above its initial height.
Momentum conservation requires the total horizontal momentum of the system to be the same before and after the bullet strikes the block. Consider the velocity before the collision to be v 0. What is the total momentum. What is the speed of the blockbullet system after the collision. Therefore the velocity of the center of mass of the system is 12 ms the velocity of the bullet block after the collision. The initial momentum of the system is simply since the block is initially at rest. From the description the collision is perfectly inelastic.
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