Physics ballistic problem please help?

Question

In a ballistics test, a 21.0 g bullet traveling horizontally at 1300 m/s goes through a 30.0 cm -thick 450 kg stationary target and emerges with a speed of 900 m/s. The target is free to slide on a smooth horizontal surface.

How long is the bullet in the target?

______s

What average force does the bullet exert on the target?

______N

What is the target's speed just after the bullet emerges?

______m/s

It is 450 kg and the values you gave me did not work. can anyone help?

Answer 1

Is the target mass 450 or 45.0 kg? I will use 450 as stated in the post, 45.0 would be a more realistic scenario, though.
Let's first calculate the speed of the target when the bullet emerges using conservation of momentum
.021*1300=450*v+.021*900
solve for v, this is the speed of the target
v=0.056/3 m/s

Now consider the bullet w/r/t the target from the moment of impact to the moment the bullet emerges from the target. This is an accelerating reference frame, so I will assume the acceleration is constant.
In this reference frame, the starting speed of the bullet is 1300 m/s, and the emerging speed of the bullet is 900-0.056/3 m/s or 899.98 m/s

the equations of motion of the bullet are
x(t)=v0*t-.5*a*t^2
and
v(t)=v0-a*t
where t is in seconds while the bullet is in the target, and a is the acceleration magnitude.
solve for t, the time the bullet is the target. With the values, the two equations are:
30=1300*t-.5*a*t^2
899.98=1300-a*t

30=1300*t-.5*400.02*t
t=0.027 seconds

now that we have t, consider the target in the Earth's frame of reference
v=a*t
a=0.684 m/s^2

And F=m*a, where F is the average force the bullet exerts on the target
308 N

j

Answer 2

The speed of the mixed mass simply after collision will also be observed utilizing inelastic collision. M×(two.15 m/s) + M×(zero m/s) = (M + M)×V V = one million.1/2 m/s The kinetic power of the mixed mass simply after collision is: Ek = m × v² / two Ek = 2M × (one million.1/2 m/s)² / two Ek = one million.1556×M m²/s² This can also be the competencies power of the mixed mass at its maximum factor, so: (one million.1556×M m²/s²) = 2M × (nine.eight m/s²) × h (zero.5778 m²/s²) = (nine.eight m/s²) × h h = zero.05896 m Now, utilizing a few trig capabilities, you'll be able to discover the highest perspective. cos(?) = (L - h) / L cos(?) = ((zero.413 m) - (zero.05896 m)) / (zero.413 m) ? = 31°

Answer 3

The above answer forgot to convert 30 centimeters into 0.3 meters when solving for the time that the bullet was in the target. Other than that, it's correct.