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 A ball of mass 160 g is thrown up at an angle of 600 to the horizontal at a speed of 10 ms-1. The angular momentum of the ball at the highest point of the trajectory  with respect to the point from which the .ball is thrown is nearly (g=10 ms-2)                

 

  • Option 1)

    1.73 kg m2/s

  • Option 2)

    3.0 kg m2/s

  • Option 3)

    3.46 kg m2/s

  • Option 4)

    6.0 kg m2/s

 

Answers (1)

best_answer

As we have learned

Angular momentum -

\vec{L}=\vec{r}\times \vec{p}

- wherein

\vec{L}  represent angular momentum of a moving particle about a point.

it can be calculated  as L=r_1\, P=r\, P_1

r_1 = Length of perpendicular on line of motion

P_1 = component of momentum along perpendicualar to r

 

 

Angular momentum about point 0 is  (mv)L_\perp \\= (mu_x)H


u_x = u/2 = 5

H = \frac{u^2 \sin ^2 \theta }{2g}= \frac{100 \times (\frac{\sqrt{3}}{2})}{2 \times 10 }= 30 /8 = 3.75 m

Angular momentum = 0.16 \times 5\times 3.75 Kgm^2 / s = 3.000 Kgm^2/s

 

 

 

 

 

 

 


Option 1)

1.73 kg m2/s

Option 2)

3.0 kg m2/s

Option 3)

3.46 kg m2/s

Option 4)

6.0 kg m2/s

Posted by

SudhirSol

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