A bullet is fired from a gun. The force on the bullet is given by F=600-(2×105)t where, F is in newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Question

A bullet is fired from a gun. The force on the bullet is given by F=600-(2×105)t where, F is in newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet? (A) 9 N -s (B) zero (C) 1.8 N-s (D) 0.9 N-s

Tardigrade Admin · Accepted Answer

When F= 0, 600 - 2×105t = 0 ∴ t=(600/2×105)=3×10-3s. Now, impulse, I=∫ limitst0 Fdt=∫ limitst0 (600-2×105t)dt 600t-2×105(t2/2)=600×3×10-3-105×(3×10-3)2 or, I= 1.8-0.9 = 0.9 N-s.

Q. A bullet is fired from a gun. The force on the bullet is given by F=600−(2×105)t where, F is in newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

9 N -s

zero

1.8 N-s

0.9 N-s

When F=0,600−2×105t=0 ∴t=2×105600​=3×10−3s. Now, impulse, I=0∫t​Fdt=0∫t​(600−2×105t)dt 600t−2×1052t2​=600×3×10−3−105×(3×10−3)2 or, I=1.8−0.9=0.9N−s.

Q. A bullet is fired from a gun. The force on the bullet is given by $F = 600 - (2 \times 1 0^{5}) t$
where, $F$ is in newton and $t$ in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?