Ablock of mass 1.3 kg is resting at the base of a frictionless ramp. a bullet of mass 50 g is traveling parallel to the ramp surface at 250 m/s. it collides with the block, enters it, and exits the other side at 100 m/s. how far up the ramp will the block travel?

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

the answer is b

explanation:

reserve is the answer