Answer:
24 J
Explanation:
given,
mass of the block 1, M = 3 Kg
initial speed in east direction, u = 5 m/s
mass of the block 2, m = 2 Kg
initial speed in west direction, u' = 2 m/s
final speed of block 1, after collision, V = 1 m/s
Kinetic energy loss during collision = ?
taking east direction positive
using conservation of momentum
M u + m u' = M V + m V'
3 x 5 + 2 x(-2) = 3 x 1 + 2 x V'
2 x V' = 8
V' = 4 m/s
loss in kinetic energy
=[tex]KE_i - KE_f[/tex]
=[tex](\dfrac{1}{2}Mu^2+\dfrac{1}{2}mu'^2) - (\dfrac{1}{2}MV^2+\dfrac{1}{2}mV'^2)[/tex]
=[tex](\dfrac{1}{2}\times 3 \times 5^2+\dfrac{1}{2}\times 2 \times (-2)^2) - (\dfrac{1}{2}\times 3 \times 1^2+\dfrac{1}{2}\times 2 \times 4^2)[/tex]
= 24 J
loss in kinetic energy is equal to 24 J
