A 10-g bullet is fired into a wooden block with a mass of 4.0 kg, giving the blo

Question

A 10-g bullet is fired into a wooden block with a mass of 4.0 kg, giving the block a velocity of 40 cm/sec. with what velocity (in m/sec) was the bullet fired?

I know the answer is 160 m/sec, but I can only get the number part of the answer I keep coming up with -160 m/sec.

The formula I am to use is

P1 + P2 = 0

m1 x v1 + m2 xv2 = 0

because the total momentum before an impact equals the total momentum after an impact.

I try to do this

v1= - (m2 x v2/m1)

but i should not have a negative answer.

Help please!!!!!!!!

ASAP

Explanation / Answer

Given that,

mass of the bullet = m1 = 10 g = 0.01 kg ; mass of the block = m2 = 4 kg ;

v(f) = 40 cm/s = 0.4 m/s

Let v1 be the initial veocity with which the bullet is fired.

Lets assume that the wooden block is at rest(being the target, and the question did not talk about its motion also).

So the total momentum of the system is due to the bullet only. from conservation of linear momentum, we know that,

P(intial) = P(final)

m1v1 + 0 = (m1+m2)vf

v1 = (m1+m2)v(f) / m1 = (0.01 + 4 ) x 0.4 / 0.01 = 160.4 m/s

Hence, the intial speed of the bullet is v1 = 160 m/s

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