Answer:
KE = 0.162 KJ
Explanation:
given,
mass of bullet (m)= 20 g = 0.02 Kg
speed of the bullet (u)= 1000 m/s
mass of block(M) = 1 Kg
velocity of bullet after collision (v)= 100 m/s
kinetic energy = ?
using conservation of momentum
m u = m v + M V
0.02 x 1000 = 0.02 x 100 + 1 x V
20 = 2 + V
V = 18 m/s
now,
Kinetic energy of the block
[tex]KE = \dfrac{1}{2}mv^2[/tex]
[tex]KE = \dfrac{1}{2}\times 1 \times 18^2[/tex]
KE = 162 J
KE = 0.162 KJ
The kinetic energy of the block is 5 J.
From the principle of conservation of linear momentum we know that momentum before collision is equal to momentum after collision. Hence;
(0.02 Kg × 1000) + (1 * 0) = (0.02 Kg × 100) + ( 1 × V)
To obtain the velocity of the block after the collision;
V = (0.02 Kg × 1000) + (1 * 0)/(0.02 Kg × 100)
V = 20/2
V = 10 m/s
The kinetic energy now is;
KE = 1/2 mv^2 = 0.5 * 1 * 10 = 5 J
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