Question

A 0.035 kg bullet strikes a 5.0 kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at 8.6 m/s. What was the original speed of the bullet?

Answer 1

QUESTION :

A 0.035 kg bullet strikes a 5.0 kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at 8.6 m/s. What was the original speed of the bullet?

Explanation:

Mass of bullet (m1) = 0.035 kg

Mass of lumber (m2) = 5.0 kg

initial velocity of lumber (u2) = 0 m/s

final velocity of lumber (v2) = 8.6 m/s

final velocity of bullet (v1) = 8.6 m/s

Now on applying the formula of conservation of momentum:-

m1u1 + m2u2 = m1v1 + m2v2

0.035*u1 + 5*0 = 0.035*8.6 + 5*8.6

0.035*u1 = 0.301 +43

u1 = 43.301 / 0.035

u1 = 1237.17 m/s

TO CHECK....

0.035*1237.17 + 0 = 43.301

43.30095 = 43.301

After rounding off.. we get

43.301 = 43.301...

Hence checked as well