Question

A stone with a mass of 0.100 kg rests on a frictionless, horizontal surface. A bullet of mass 2.40 g traveling horizontally at 525 m/s strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 315 m/s .A stone with a mass of 0.100 kg rests on a frictionless, horizontal surface. A bullet of mass 2.40 g traveling horizontally at 525 m/s strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 315 m/s . A) Compute the magnitude of the velocity of the stone after it is struck.
B) Compute the direction of the velocity of the stone after it is struck. (degrees from the initial direction of the bullet)

Answer 1

Answer:

a. the magnitude of the speed is 14.69 m/s

b. The direction is - 30 ° 57' 49''

Explanation:

As there are no external forces in the horizontal direction, we know that the linear momentum must be conserved.

$\vec{p} = constant$

So, taking $\vec{p}_i$ our initial momentum and  $\vec{p}_f$ our final momentum, we got:

$\vec{p}_i=\vec{p}_f$

The momentum of the system will be:

$\vec{p} = m_b \ \vec{v}_b + m_s \ \vec{v}_s$

where $m_b$ and $\vec{v}_b$ are the mass and the velocity of the bullet and $m_s$ and $\vec{v}_s$ are the mass and the velocity of the stone.

Initial conditions

We know that the mass of the bullet is :

$m_b = 2.4  \ g = 2.4 \ 10^{-3} \ kg$

taking the unit vector $\hat{i}$ pointing in the original direction of the bullet, the velocity will be:

$\vec{v}_{b_i} =  525 \frac{m}{s} \ \hat{i}$

The mass of the stone is

$m_{s} = 0.100 \  kg$

as is at rest, the initial velocity will be zero

$\vec{v}_{s_i} =  0$

So, our initial momentum will be

$\vec{p}_i = m_b \ \vec{v}_{b_i} + m_s \ \vec{v}_{s_i}$

$\vec{p}_i = 2.4 \ 10^{-3} \ kg \ * 525 \frac{m}{s} \ \hat{i}+  0.100 \  kg \ *  0$

$\vec{p}_i = 1.26 \frac{kg \ m}{s} \ \hat{i}$

Final conditions

The masses will be the same.

Taking the unit vector $\hat{j}$ pointing in the final direction of the bullet, the velocity of the bullet is

$\vec{v}_{b_f} =  315 \frac{m}{s} \ \hat{i}$

So, the final momentum will be

$\vec{p}_f = 2.4 \ 10^{-3} \ kg \ * 315 \frac{m}{s} \ \hat{j}+  0.100 \  kg \ *  \vec{v}_{s_f}$

$\vec{p}_f = 0.756 \frac{kg \ m}{s} \ \hat{j}+  0.100 \  kg \ *  \vec{v}_{s_f}$

Obtaining the velocity

Now, we use

$\vec{p}_i=\vec{p}_f$

and we obtain

$1.26 \frac{kg \ m}{s} \ \hat{i} = 0.756 \frac{kg \ m}{s} \ \hat{j}+  0.100 \  kg \ *  \vec{v}_{s_f}$

working it a little

$ 0.100 \  kg \ *  \vec{v}_{s_f} = 1.26 \frac{kg \ m}{s} \ \hat{i} - 0.756 \frac{kg \ m}{s} \ \hat{j}$

$ \vec{v}_{s_f} = \frac{(1.26 \frac{kg \ m}{s} , - 0.756 \frac{kg \ m}{s}) }{ 0.100 \ kg}$

$ \vec{v}_{s_f} = (12.6 \frac{m}{s} , - 7.56 \frac{m}{s})$

This is the final velocity of the stone. Now, we can obtain the magnitude using the Pythagorean theorem:

$|\vec{v}| = \sqrt{{v}_{x} ^2 + {v}_{y}^2}$

$|\vec{v}_{s_f}| = \sqrt{(12.6 \frac{m}{s}) ^2 + (- 7.56 \frac{m}{s})^2}$

$|\vec{v}_{s_f}| = 14.69 \ \frac{m}{s}$

We can obtain the angle as:

$\theta = arctan(\frac{v_y}{v_x})$

$\theta = arctan(\frac{ - 7.56 \frac{m}{s} }{ 12.6 \frac{m}{s} })$

$\theta = - 30 \° 57' 49''$