Acannon of 2000 kg fires a shell of 10 kg at
a speed of 100 m/s. what is the velocity
of the cannon? if the cannon slowed down
with an acceleration of -0.25 m s-2, what
is the minimum safe distance you should
stand behind it?

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

After the shot, the momentum is:

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

And solving for V, we find the velocity of the cannon:

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

a, c, d, e are all applying about success strategies.