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A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sandbag,how deep,in meters,did the bullet penetrate the sandbag

User Amaury
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Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is
v_i=180~m/s, the final velocity is
v_i=0~m/s, and the total time of the motion is
\Delta t=0.02~s, so the acceleration is given by

a= (v_f-v_i)/(\Delta t) = -9000~m/s^2
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using

S=v_i t + (1)/(2)at^2 = 180~m/s \cdot 0.02~s + (1)/(2)(-9000~m/s^2)(0.02~s)^2=1.8~m
So, the bullet penetrates the sandbag 1.8 meters.
User Birko
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