Answer:
s = 2.65 m
Explanation:
given,
mass of block,M = 1.3 Kg
mass of bullet,m = 50 g = 0.05 Kg
speed of bullet,u = 250 m/s
speed of bullet after collision,v = 100 m/s
distance traveled by the block = ?
Assuming the angle of inclination of ramp equal to 40°
calculating the speed of the block
using conservation of momentum
M u' + m u = m v + M v'
initial speed of the block is equal to zero
0 + 0.05 Kg x 250 = 0.05 x 100 + 1.3 x v'
1.3 v' = 7.5
v' = 5.77 m/s
now, calculation of acceleration
equation the horizontal component
-mg sin θ = ma
a = - g sin θ
a = - 9.8 x sin 40°
a = -6.29 m/s²
using equation of motion for the calculation of distance moved by the block
v² = u² + 2 a s
0² = 5.77² + 2 x (-6.29) x s
12.58 s = 33.29
s = 2.65 m
hence, the distance moved by the block is equal to 2.65 m
