CP? A .22-caliber rifle bullet traveling at 350 m/s strikes a large tree and penetrates it to a depth of 0.130 m. The mass of the bullet is 1.80 g. Assume a constant retarding force. (a) How much time is required for the bullet to stop? (b) What force, in newtons, does the tree exert on the bullet?

Solution 30E Step 1 of 5: Given data, Mass of the bullet, m=1.8 g Using 1g = 10 kg m = 0.0018 kg Initial speed of bullet, u= 350 m/s Penetrating depth is the distance, s= 0.13 m Since the bullet should stop at the penetrating depth, the final velocity will be zero, v=0 As shown in the figure below, Step 2 of 5: Assume a constant retarding force. (a) How much time is required for the bullet to stop Using kinematic equation, v= u +at Solving for time, t= vu…………...1 a In order to calculate the time taken, we need to calculate acceleration of the bullet. Step 3 of 5: To calculate the acceleration of the bullet, Using kinematic equation, v = u + 2as Solving for acceleration, v u2 a = 2s Substituting v= 0 , u = 350 m/s and s=0.13m 2 a = 0 (350 m/s) 2(0.13m) 2 a=471153 m/s Acceleration of the bullet in magnitude is found to be 471153 m/s . 2