Bullet shot into a block
A 21 g bullet is fired upwards 100 ft under a 1.40 kg block of wood. If the bullet is initially shot at a speed of 300 m/s , how high will the block rise in the air after the bullet embeds into it?
Do I just do conservation of momentum from the 300 m/s bullet to the bullet and block together?
Answer
Be careful, because this problem has multiple steps. First, you have a kinematic problem to determine the bullet's speed right before it collides with the block. Then you have the collision itself in which momentum must be conserved. Then you have another kinematic problem where the block and bullet combination rises into the air and reaches its peak.
Make sure you treat each part of the problem separately and check your results before moving on to make sure they make sense. The bullet should lose a little bit of speed before it collides with the block. The momentum of the bullet before the collision is equal to the momentum of the block and bullet combination, which now has a combined mass that is much greater. Then the velocity of the block and bullet combination after the collision acts as your initial velocity for the final kinematic stage.
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Be careful, because this problem has multiple steps. First, you have a kinematic problem to determine the bullet's speed right before it collides with the block. Then you have the collision itself in which momentum must be conserved. Then you have another kinematic problem where the block and bullet combination rises into the air and reaches its peak.
Make sure you treat each part of the problem separately and check your results before moving on to make sure they make sense. The bullet should lose a little bit of speed before it collides with the block. The momentum of the bullet before the collision is equal to the momentum of the block and bullet combination, which now has a combined mass that is much greater. Then the velocity of the block and bullet combination after the collision acts as your initial velocity for the final kinematic stage.