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Rudiy27
4 years ago
14

A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t

o each other, at 24 m/s . the third piece has twice the mass as the other two. you may want to review (page 270) . part a what is the speed of the third piece?

Physics
2 answers:
netineya [11]4 years ago
8 0
Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
m_{1}  v_{1i} + m_{2}  v_{2i} + m_{3}  v_{3i} = m_{1}  v_{1f} + m_{2}  v_{2f} + m_{3}  v_{3f}
Before the firecracker blows a coconut does not move, so left side is equal to 0:
0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
v_{3f} = -  \frac{m_{1}  v_{1f}  + m_{2} v_{2f} }{ m_{3} }

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s
This speed we can insert into formula for v3f:
v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\  \\ v_{3f} = - \frac{2m*33.9 }{ 2m }  \\  \\ v_{3f} = - 33.9m/s

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.
expeople1 [14]4 years ago
4 0

The speed of the third piece is about 17 m/s

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

mass of object 1 = m₁ = m

mass of object 2 = m₂ = m

velocity of object 1 = v₁ = -24 j

velocity of object 2 = v₂ = -24 i

mass of object 3 = m₃ = 2m

<u>Asked:</u>

velocity of object 3 = v₃ = ?

<u>Solution:</u>

We will use conservation of momentum formula to solve this problem:

m_1u_1 + m_2u_2 + m_3u_3 = m_1v_1 + m_2v_2 + m_3v_3

0 + 0 + 0 = m(-24 \widehat{j}) + m (-24 \widehat{i}) + 2mv_3

0 = -24 \widehat{j} -24 \widehat{i} + 2v_3

2v_3 = 24 \widehat{i} + 24 \widehat{j}

v_3 = (24 \widehat{i} + 24 \widehat{j}) \div 2

\large {\boxed {v_3 = 12 \widehat{i} + 12 \widehat{j}} }

\texttt{ }

<em>We could calculate the magnitude of the vector of velocity of this third piece as follows:</em>

|v_3| = \sqrt{12^2 + 12^2}

|v_3| = \sqrt{2 (12^2)}

|v_3| = 12\sqrt{2} \texttt{ m/s}

|v_3| \approx 17 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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