A 5 g bullet moving with an initial speed of 530 m/s is fired into and passes th

Question

A 5 g bullet moving with an initial speed of 530 m/s is fired into and passes through a 2 kg block. the block, initially at rest, on a frictionless horizontal surface, is connected to a spring with force constant 800 N/m. the block moves 6 cm to the right after impact before being brought to rest by the spring.

a) find the speed of the block just after the impact.

b) find the speed at which the bullet emerges from the block.

c) what quantity is conserved in this collision? is mechanical energy of the bullet-block-system conserved during the collision?

Explanation / Answer

Momentum before Pb = 0.005*500 = 2.5kgm/s
The force that depresses the spring will change the momentum of the block
which had it's momentum changed by the impulse of the bullet
F = kx = 0.013*444 = 5.772N
a = F/m = 5.772/5 = 1.1544m/s^2 and is deceleration
Now use Vf^2 = Vi^2 + 2*a*d = 0
Vi = sqrt[2*1.1544*0.013] = 0.03m/s
So the momentum of the block was 0.03*5 =0.15kgm/s
Thus the momentum of the bullet will 2.5 - 0.15 = 2.35kgm/s
2.35 = 0.005*V => V = 470m/s <-------------------- A
0.0025*470^2 = 552.2J
0.0025*500^2 = 625 J
625-552.2 = 72.8J <------------------- B

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.