Answer:
v₁ = 4 [m/s].
Explanation:
This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]

where:
m₁ = mass of the tank = 500 [kg]
v₁ = velocity of the tank after firing the missile [m/s]
m₂ = mass of the missile = 20 [kg]
v₂ = velocity of the missile after firing = 100 [m/s]
![(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]](https://tex.z-dn.net/?f=%28500%2Av_%7B1%7D%29%3D%2820%2A100%29%5C%5Cv_%7B1%7D%3D2000%2F500%5C%5Cv_%7B1%7D%3D4%5Bm%2Fs%5D)
Complete Question
The diagram for this question is showed on the first uploaded image (reference homework solutions )
Answer:
The velocity at the bottom is 
Explanation:
From the question we are told that
The total distance traveled is 
The mass of the block is 
The height of the block from the ground is h = 0.60 m
According the law of energy

Where PE is the potential energy which is mathematically represented as

substituting values


So
KE is the kinetic energy at the bottom which is mathematically represented as

So

substituting values
=> 
=> 
=> 
A beta particle is identical to : an electron
Both of beta particle and electron are high in energy and move in high speed.
hope this helps