Question

An explosion breaks a 26.0-kg object into three parts. The object is initially moving at a velocity of 30.0 m/s due west. Part (1) has a mass m 1 = 4.90-kg and a velocity of 58.0 m/s due west. Part (2) has a mass m 2 = 6.20-kg and a velocity of 72.0 m/s due south. What is the magnitude of the velocity of part (3)?

Answer 1

Answer:

The magnitude of the velocity of part 3 is 46.60 m/s

Explanation:

We can solve this problem using conservation of linear momentum.

As there are no external forces in the horizontal direction, we know

$\frac{\vec{p}_{hor}}{dt}= \vec{F}_{hor}=0$

$\vec{p}_{hor}=const$

So, the horizontal linear momentum is conserved.

Using the unit vector $\hat{i}$ pointing west and the unit vector $\hat{j}$ pointing south,  the initial momentum is

$\vec{p}_i=m \ \vec{v}_i = 26 \kg \ 30 \frac{m}{s} \hat{i}= 780 \frac{kg \ m}{s} \hat{i}$

and the final momentum is

$\vec{p}_f=m_1 \ \vec{v}_1 + m_2 \ \vec{v}_2+m_3 \ \vec{v}_3$

with velocities:

$\vec{v}_1 = 58 \frac{m}{s} \hat{i}$

$\vec{v}_1 = 72 \frac{m}{s} \hat{j}$

And, as the total mass is :

$m_{total} = m_1 + m_2 + m_3$

so, the third mass is :

$m_3 = m_{total} - m_1 - m_2$

$m_3 = 26 \ kg - 4.90 \ kg - 6.20 \ kg$

$m_3 = 14.90 \ kg$

$\vec{p}_f=4.90 \ kg \ 58 \frac{m}{s} \hat{i} + 6.20 \ kg \ 72 \frac{m}{s} \hat{j} + 14.90 \ kg \ \vec{v}_3$

$\vec{p}_f=248.20 \frac{kg \ m}{s} \hat{i} + 446.40 \frac{kg \ m}{s} \hat{j} + 14.90 \ kg \ \vec{v}_3$

but this must be equal to the initial momentum

$780 \frac{kg \ m}{s} \hat{i}=248.20 \frac{kg \ m}{s} \hat{i} + 446.40 \frac{kg \ m}{s} \hat{j} + 14.90 \ kg \ \vec{v}_3$

so:

$14.90 \ kg \ \vec{v}_3 = 780 \frac{kg \ m}{s} \hat{i}-248.20 \frac{kg \ m}{s} \hat{i} - 446.40 \frac{kg \ m}{s} \hat{j}$

$14.90 \ kg \ \vec{v}_3 = 531.8 \frac{kg \ m}{s} \hat{i} - 446.40 \frac{kg \ m}{s} \hat{j}$

$\vec{v}_3 = \frac{ 531.8 \frac{kg \ m}{s} \hat{i} - 446.40 \frac{kg \ m}{s} \hat{j} }{14.90 \ kg }$

$\vec{v}_3 = 35.69 \frac{ m}{s} \hat{i} - 29.97 \frac{ m}{s} \hat{j}$

To find the magnitude, we just need to use the Pythagorean theorem

$|\vec{v}_3| = \sqrt{ (v_{3_x} )^2 + (v_{3_y} )^2 }$

$|\vec{v}_3| = \sqrt{ (35.69 \frac{ m}{s} )^2 + (29.97 \frac{ m}{s} )^2 }$

$|\vec{v}_3| =  46.60 \frac{ m}{s} }$

Answer 2

Velocity is defined as the rate of change of displacement of the body. The magnitude of the velocity of the part (3) will be 46.60 m/sec.

What is the law of conservation of linear momentum?

According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state that about how momentum gets conserved.

The given data in  the problem is

Mass of object before explosion= m

v is the initial velocity of mass m = 30.0 m/sec

mass of first part m₁= 4.90 kg

mass of second part m₂=6.20 kg

mass of third part m₃ =?

v₁ is the velocity of mass m₁ in the west=58.0 m/sec.

v₂ is the velocity of the mass m₂= 72.0 m/sec.

v₃  is the velocity of the mass m₃=?

By unit vector indicating west and the unit vector  indicating south,  the initial momentum is;

The value of mass 3 is found by

m=m₁+m₂+m₃

m₃=m-m₁-m₂

m₃= 26.0-4.90-6.20

m₃=14.90 Kg

The initial momentum of the body will be

$\rm p_i=mv_i\\\\\rm p_i=26\times30\; \vec i\\\\\rm p_i=780 Kgm/sec.$

$\rm {p_F=m_1v_1+m_2v_2+m_3v_3}$

$\vec v_1=58.0 m/sec\\\vec v_2 = 72 m/sec$

The value of final momentum

$\rm {p_F=m_1v_1 \vec i+m_2v_2 \vec j+m_3v_3}\\\\\rm {p_F=4.90 \times 58 \vec i +6.20 \times 72 \vec j+14.90 \times v_3}\\\\\rm {p_F=248.20 \vec i+446.40 \vec j+14.90v_3}\\\\\\\\ \vec v_3=(35.69 \vec i-29.97 \vec\;j)m/sec$

The resultant value of velocity 3 is obtained by;

$\rm v_3=\sqrt{(v_3)^2_x +( v_3)^2_y} \\\\v_3=\sqrt{(35.69)_x +( 29.97)_y}\\\\v_3=46.60 m/sec$

Hence the magnitude of the velocity of the part (3) will be 46.60 m/sec.

Learn more about the law of conservation of momentum refer to link;

https://brainly.com/question/1113396