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

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.