A 20-g bullet is shot vertically into an 2.8-kg block. The block lifts upward 9 mm. The bullet penetrates the block and comes to rest in it in a time interval of 5 ms. Assume the force on the bullet is constant during penetration and that air resistance is negligible. What is the speed of the bullet just before the impact

The speed of the bullet just before the impact is 701 m/s

Explanation:

Given;

mass of the bullet, m₁ = 20 g = 0.02 kg

mass of the block, m₂ = 2.8 kg

displacement of the block, d = 9 mm = 9 x 10⁻³ m

duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s

Apply the principle of conservation of energy;

The final kinetic energy of the bullet = maximum potential energy of the block

Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.

m₁u₁  +  m₂u₂  =  v(m₁  +  m₂)

where;

u₁ is the initial velocity of the bullet

u₂ is the initial velocity of the block = 0

m₁u₁  +  0  = v(m₁  +  m₂)

m₁u₁  =  v(m₁  +  m₂)

0.02u₁ = 4.97(2.8 + 0.02)

0.02u₁ = 14.02

u₁ = 14.02 / 0.02

u₁ = 701 m/s

Therefore, the speed of the bullet just before the impact is 701 m/s

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

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