Question

Two billiard balls are moving on a table and the component velocities along the length and
breadth are 5,5 ms^-1 for one ball 2√3,2ms^-1 for the other ball the angle between the motion of balls
1) 30°
2) 60°
3)40°
4)15°​

Answer 1

Answer:

4)15° IS THE RIGHT ANSWER

Explanation:FOR THE FIRST BALL VX=1m/S AND VY=?3

?=60

FOR THE SECOND BALL,TAN?=V?Y/V?X=2/2=1

?=45°

 HENCE,THE ANGLE BETWEEN THE PATHOF TWO BALLS =60°-45°=15°

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Answer 2

Answer:

The angle between the motion of balls is 15°.

Explanation:

Given the velocities of two balls: $\vec v_1= 5\hat i+5\hat j ~ms^{-1}~ \mbox {and} ~\vec v_2= 2\sqrt{3} \hat i+2\hat j~ms^{-1}$

The angle between two vectors is given by

$cos\theta=\dfrac{\vec A.\vec B}{|\vec A||\vec B|}$

Substituting the given values, we get

$cos\theta=\dfrac{\vec v_1.\vec v_1}{|\vec v_1|.|\vec v_1|}$

      $=\dfrac{(5\hat i+5\hat j).(2\sqrt{3} \hat i+2\hat j)}{\sqrt{5^{2} +5^{2} }.\sqrt{(2\sqrt{3} )^{2} +2^{2} } }$

      $=\dfrac{5\hat i.2\sqrt{3} \hat i+5\hat j.2\hat j}{\sqrt{25 +25 }.\sqrt{12+4 } }$

      $=\dfrac{5.2\sqrt{3} +5.2}{5\sqrt{2}\times 4 }=\dfrac{5.2(\sqrt{3} +1)}{5\sqrt{2}\times 4 }$

      $=\dfrac{\sqrt{3} +1}{2\sqrt{2} }$

As we know from the trigonometric ratios,

$cos 15^o=cos (45^o - 30^o)$

          $= cos 45^o. 30^o + sin 45^o. sin 30^o$

         $= \frac{1}{\sqrt{2} } \frac{\sqrt{3} }{2 } + \frac{1}{\sqrt{2} } \frac{1}{2}$

         $= \frac{\sqrt{3}+1 }{2\sqrt{2} }$

Therefore, from the above result,

$cos \theta=\dfrac{\sqrt{3} +1}{2\sqrt{2} }$ we get $\theta = 15^{o}$

Therefore, the angle between the motion of balls is 15°.