Question
A small wooden block with mass 0.800kg is suspended from the lower end of a light cord that is 1.52m long. The block is initially at rest. A bullet with mass 0.0126kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.775m , the tension in the cord is 4.82N. What was the initial speed v0 of the bullet?
Answers
bobpursley
tension=mg*cosTheta+v^2/r
You know cosTheta, it is (1.52-.775)/1.52
solve for v at that point.
Now figure total energy of the block bullet...you know KE+ PE
totalENergy=1/2 m v^2+ m g h
that energy is the same energy of the block when it absorved the bullet...
solve for Velocity of the block/bullet then from total energy when h=0
Now, using conservation of momentum,
momentumbullet= momentumBlockBullet
now solve for velocity of the bullet before impace.
You know cosTheta, it is (1.52-.775)/1.52
solve for v at that point.
Now figure total energy of the block bullet...you know KE+ PE
totalENergy=1/2 m v^2+ m g h
that energy is the same energy of the block when it absorved the bullet...
solve for Velocity of the block/bullet then from total energy when h=0
Now, using conservation of momentum,
momentumbullet= momentumBlockBullet
now solve for velocity of the bullet before impace.
No
267.96m/s
sug mah bawls
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