A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?

mannyell

2 years ago

See how i got mine ooh cuz you people are confusing me 🤲
m1u1 + m2u2 = (m1 + m2 )v
(0.05 x 200) + (0.95 x 0) = (0.05 + 0.95)v
(10) +(0) =1v
v=10/1 =10
kinetic energy after impact =½mv² + ½mv²
½(0.05)(10)² + ½(0.95)(10)²
=(2.5)+(47.5)
=50J
Ooh it was actually the same thing but I've typed a ton ‍‍

Ire01

4 years ago

Can someone pls explain better

ogunleyeferanmi5

4 years ago

I don't even understand this 1 ☝️

De sage

1 year ago

use this formula
M1U1 +0= V(M1+M2)

then use
KE2= 1/2{M1+M2}V²

Odemosad

1 year ago

m1u1 + m2u2 = v(m1 + m2) [law of conversation of momentum]
NOTE: m2u2 = 0 (since block is at rest)
and v is common.
Hence, v = 10 m/s

But K.E = ½mv² = ½ x 1 x 100 = 50J