A 1 .50 times 10^-2 Kg bullet travelling at 5.00 times 10^2 m/s strikes an 0.800
ID: 1487622 • Letter: A
Question
A 1 .50 times 10^-2 Kg bullet travelling at 5.00 times 10^2 m/s strikes an 0.800 kg block of wood The collusion between the bullet and the block of wood lasts seconds The block of wood rests initially on a table edge 0 800 m above the ground. The bullet buried itself in the block. Find the velocity of the bullet-block system after the collision. Find the magnitude of the impulse on the block due to the collision. Rno the average force acting on the block from the bullet. Find the magnitude of the velocity of .thee center of mass of the bullet-block system before the collision.Explanation / Answer
(a)
apply the law of conservation of momentum, we get
m1u1+m2u2 = [m1+m2]v
the common speed of the system is,
v = m1u1/m1+m2 = 0.015*500 / 0.015+0.8 = 9.2 m/s
---------------------------------------------------------------------------------------
the final velcoity of the blok when it strikes the ground is,
V = sqrt[2*gh] = sqrt[2*9.8*0.8] = 3.96 m/s
the impulse is,
I = M [v-V] = [0.015+0.8][9.2-3.96] = 4.27 N.s
the average force is,
F = dp/dt = 4.27 N.s/1.5x10-3 = 2847.176 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.