Answer:
v' = 1.5 m/s
Explanation:
given,
mass of the bullet, m = 10 g
initial speed of the bullet, v = 300 m/s
final speed of the bullet after collision, v' = 300/2 = 150 m/s
Mass of the block, M = 1 Kg
initial speed of the block, u = 0 m/s
velocity of the block after collision, u' = ?
using conservation of momentum
m v + Mu = m v' + M u'
0.01 x 300 + 0 = 0.01 x 150 + 1 x v'
v' = 0.01 x 150
v' = 1.5 m/s
Speed of the block after collision is equal to v' = 1.5 m/s
It’s easier to lift the wheelbarrow when the load is near the front because there isn’t any pressure near the back where the handles are.
Farmula 2gh=vf²-vi²
2×0.42×9.81=0²-vi²
8.2404=vi²
Vi=✓8.2404
Vi=2.87m/sec
