What is the speed and mass of the object?

The recoil of a projectile launcher finds that the linear momentum of an ejected object is 26.1 kg m/s, and the target is equipped with an energy detector that determines that the object has a kinetic energy of 200 J.

2 Answers
Feb 28, 2018

speed = 15.3256705$\frac{m}{s}$
mass = 1.703025 $k g$

Explanation:

From the Kinetic Energy and momentum formulas
$K . E = \frac{1}{2} \cdot m \cdot {v}^{2}$

and momentum
$P = m v$

we can get
$K . E = \frac{1}{2} \cdot P \cdot v$

and we can get
$K . E = {P}^{2} / \left(2 m\right)$
because $v = \frac{P}{m}$

so

for the speed, I will use $K . E = \frac{1}{2} \cdot P \cdot v$

$200 J = \frac{1}{2} \cdot 26.1 k g \frac{m}{s} \cdot v$
$V = \frac{200 J}{\left(26.1 k g \frac{m}{s}\right) \cdot \frac{1}{2}} = 15.3256705 \frac{m}{s}$

for the mass, I will use $K . E = {P}^{2} / \left(2 m\right)$
$m = {P}^{2} / \left(2 K . E\right)$
$m = \frac{{26.1}^{2} k g \frac{m}{s}}{2 \cdot 200 J}$ = 1.703025kg

Feb 28, 2018

$m = 1.70 k g$

$v = 15.32 \frac{m}{s}$ away from the launcher.

By solving a system of equations.

Explanation:

We know the following based on the equations for momentum and kinetic energy.

$m \cdot v = 26.1 \frac{k g m}{s}$ (this is equation1)

$\frac{1}{2} m \cdot {v}^{2} = 200 J$ (this is equation2)

To solve the above system of equations, we need to isolate a variable. Let us first isolate mass to solve for velocity.

$m = \frac{26.1}{v}$ (this is equation1)

$m = \frac{400}{v} ^ 2$ (this is equation2)

And because mass is equal we can combine the equations to solve for v.

$\frac{26.1}{v} = \frac{400}{v} ^ 2$

$v = \frac{400}{26.1}$

$v = 15.32 \frac{m}{s}$ away from the launcher.

Finally, we can solve for mass by plugging our velocity back in to the momentum equation You could also find it other ways too.

$p = m \cdot v$

$26.1 = m \cdot 15.32$

$m = 1.70 k g$