9.47. Model: Let the system be bullet + target. No external horizontal forces act on this system, so the
horizontal momentum is conserved. Model the bullet and the target as particles. Since the target is much more
massive than the bullet, it is reasonable to assume that the target undergoes no significant motion during the brief
interval in which the bullet passes through.
Visualize:
Solve: (a) By assuming that the target has negligible motion during the interval in which the bullet passes
through, the time is that needed to slow from 1200 m/s to 900 m/s in a distance of 30 cm. We’ll use kinematics to
first find the acceleration, then the time.
(v1x ) B2 = (v0 x ) B2 + 2ax Δx
⇒ ax =
⇒ Δt =
(v1x ) B2 − (v0 x ) B2 (900 m/s)2 − (1200 m/s)2
=
= −1.05 × 106 m/s 2
2Δx
2(0.30 m)
(v1x ) B = (v0 x ) B + ax Δt
(v1x ) B − (v0 x ) B 900 m/s − 1200 m/s
=
= 2.86 × 10−4 s = 286 μ s
ax
−1.05 × 106 m/s 2
The average force on the bullet is Favg = m ax = 26 kN.
(b) Now we can use the conservation of momentum equation p1x = p0 x to find
mT (v1x )T + mB (v1x ) B = mT (v0 x )T + mB (v0 x ) B = 0 + mB (v0 x ) B
⇒ (v1x )T =
mB (v0 x ) B − mB (v1x ) B 0.025 kg
=
( (1200 m/s) − (900 m/s) ) = 0.021 m/s
mT
350 kg