A lead bullet of mass 0.05 kg is fired with a velocity of 200 m/s into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is

JThomson

7 years ago

0.05×200=(0.05+0.95)v
v=10m/s
Bullet is stuck in lead block; thus total mass is (0.95+0.05)=1kg
Kinetic energy (KE)=1/2×(0.95+0.05)×10^2
=1/2×1×100
=50J

jecyemma

7 years ago

Applying principles of moment
m1u=(m1+m2)V
:.V=m1u/m1+m2
V=0.5×200/0.5+0.95
V=10/1
V=10m|s
Kinetic energy = 1/2V^²(m1+m2)
1/2×(10)²(0.05+0095)
1/2×100(1)
=50j

Brainy55

6 years ago

Pls I need the answer with workings

lekestephen

2 years ago

Workings not correct oo

Sensei

2 months ago

Correct 💯

Littleleo

7 years ago

mV=Mv (newton third law of motion)
m=0.05kg
V=200ms-1 ,v=?
M=0.95kg
0.05*200=0.95*v
v=10/0.95=10.53
ke=1/2mv^2 =1/2*0.95*10.53*10.53=52.6~50j

Fmpunky

5 years ago

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