A 2.9 gram bullet is shot into a tree stump. It enters at a velocity of 304 m/sec and comes to rest after having penetrated the stump in a straight line. It takes 0.00043 sec to come to a stop after striking the stump. What was the force on the bullet during impact?

Question

Rose James

Physics

2 February, 07:07

A 2.9 gram bullet is shot into a tree stump. It enters at a velocity of 304 m/sec and comes to rest after having penetrated the stump in a straight line. It takes 0.00043 sec to come to a stop after striking the stump. What was the force on the bullet during impact?

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Livia Shah · Answer 1

First convert grams to kilograms (so that the answer will be N) 1000g=1kg so 2.9g=.0029kg.

To solve for force we use the equation F=ma, so we need to find acceleration (a). Next use the equation a=Δv/Δt. Δt is given as being. 00043 seconds. Δv is the difference in velocities - final minus initial (this is important to discern the direction of the force vector); 0-304=-304. Now plugin and solve for a. The result is - 706976.7442 m/s², or - 710000 m/s² if you are using significant figures.

Now use the values for mass and acceleration to solve for force. F=ma=-2050.232558N or F = - 2100N (if you are using significant figures).