So here's some questions on Laws of Motion which appears to be tough but are very simple :
First we will sort out what information we are given
Mass of bullet, m=0.1 kg
Initial velocity of bullet, u=20 ms^-1
Penitrated length, d=0.2 m
Now, v^2 - u^2 = 2as
Since bullet stops after penitrating so, v=0 ms^-1
=> -(20)^2 = -2a*0.2 (since we are finding resistence the sign of 'a' is as -ve)
=>200 = 0.2a
=> a = 1000ms^-2
So, resistance is F ofered by wood = ma
=> F = 0.1x1000 = 100 N
So, force F=50 N
Mass m=2 kg
Force is acting at an angle of 30° with the vertical axis so, with horizontal axis the force is acting at an angle of 60°.
=>Fcos60° = ma
=> 50x(1/2) = 2a
=> a = 12.5 ms^-2
Mass of each bullet = 0.03 kg
Fire rate = 400 per second
The total mass released by mashine gun per second, m = 0.3*400 = 12 kg
Velocity of each bullet, v = 400 ms^-1
Since, we have to make an equilibrium we have to give same amount of force of that of gun
=> F = 12x400 = 4800 N
1:
Mass of car, m1=1000 kg
Initial speed of car, u=30 ms^-1
Mass of stationary lorry, m2=9000 kg
Final velocity of both = V
By Conservation of Momentum
m1u = MV ( M = m1 + m2 since they jammed together )
=>1000x30 = 10000xV
=>V = 3 ms^-1
2:
Mass of bullet, m1 = 0.007 kg
Mass of metal block = 7 kg
Final velocity of both = 70 cms^-1 = 0.7 ms^-1
Initial velocity of bullet = U ms^-1
By Conservation of Momentum
m1u = MV
=> 0.007xU = 7.007x0.7
=> U = 700.7 ms^-1