Physics, 17.01.2020 02:31 knighiv

A200 kg stationary cannon fires a 12 kg cannonball to the right at 300 m/s. explain. a) calculate the momentum of the cannon before the collision. b) calculate momentum of the cannonball before the collision. c) calculate the total kinetic energy before the collision. d) calculate the momentum of the cannon after the collision.

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a) Since before collision the velocity of the cannon is zero and we know that momentum is the product of mass and velocity so its momentum is also zero.

b) Since before collision the velocity of the cannonball is zero and we know that momentum is the product of mass and velocity so its momentum is also zero.

c) $\therefore KE=0$

d) $p=3600\ kg.m.s^{-1}$

Explanation:

Given:

mass of the cannon, $M=200\ kg$

mass of cannonball fired, $m=12\ kg$

velocity of cannonball, $v=300\ m.s^{-1}$

Momentum of the cannon before collision:

Since before collision the velocity of the cannon is zero and we know that momentum is the product of mass and velocity so its momentum is also zero.

Since before collision the velocity of the cannonball is zero and we know that momentum is the product of mass and velocity so its momentum is also zero.

The velocity before collision is zero.

kinetic energy is given as:

$KE=\frac{1}{2} m.v^2$

$\therefore KE=0$

Momentum of cannon after collision:

p=M.v'

considering the collision to be elastic, its momentum will be equally opposite of the cannonball:

p=m.v=M.v'

$p=12\times300$

$p=3600\ kg.m.s^{-1}$

Answer from: Quest

your answer is c! hope this : )

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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