Home
Class 12
PHYSICS
A body is projected at an angle 60^(@) w...

A body is projected at an angle `60^(@)` with the horizontal ground with kinetic energy k. When the velocity makes an angle `30^(@)` with the horizontal, the kinetic energy of the body will be `(k)/(a)`. Find value of a.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Understand the initial conditions A body is projected at an angle of \(60^\circ\) with the horizontal with an initial kinetic energy \(K\). The initial velocity can be expressed as: \[ K = \frac{1}{2} m u^2 \] where \(u\) is the initial speed of the body and \(m\) is its mass. ### Step 2: Analyze the velocity components When the body reaches a point where the velocity makes an angle of \(30^\circ\) with the horizontal, we need to find the new velocity \(v\) at that point. The horizontal component of the velocity remains constant throughout the projectile motion. The horizontal component of the initial velocity is: \[ u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \] At the point where the angle is \(30^\circ\), the horizontal component of the velocity is: \[ v_x = v \cos(30^\circ) = v \cdot \frac{\sqrt{3}}{2} \] ### Step 3: Set the horizontal components equal Since the horizontal component of velocity remains constant, we can set the two expressions for horizontal velocity equal to each other: \[ \frac{u}{2} = v \cdot \frac{\sqrt{3}}{2} \] ### Step 4: Solve for \(v\) Rearranging the equation gives: \[ v = \frac{u}{\sqrt{3}} \] ### Step 5: Calculate the new kinetic energy The kinetic energy \(K'\) at the point where the velocity makes an angle of \(30^\circ\) is given by: \[ K' = \frac{1}{2} m v^2 \] Substituting \(v = \frac{u}{\sqrt{3}}\): \[ K' = \frac{1}{2} m \left(\frac{u}{\sqrt{3}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{3} = \frac{1}{3} \left(\frac{1}{2} m u^2\right) = \frac{K}{3} \] ### Step 6: Relate \(K'\) to \(K\) We know that \(K' = \frac{K}{a}\). From our previous result, we have: \[ \frac{K}{3} = \frac{K}{a} \] ### Step 7: Solve for \(a\) By comparing both sides, we find: \[ a = 3 \] ### Final Answer The value of \(a\) is \(3\). ---

To solve the problem, we will follow these steps: ### Step 1: Understand the initial conditions A body is projected at an angle of \(60^\circ\) with the horizontal with an initial kinetic energy \(K\). The initial velocity can be expressed as: \[ K = \frac{1}{2} m u^2 \] where \(u\) is the initial speed of the body and \(m\) is its mass. ...
Promotional Banner

Topper's Solved these Questions

  • REVIEW TEST

    VIBRANT|Exercise PART - II : PHYSICS|262 Videos
  • TEST PAPERS

    VIBRANT|Exercise PART - II : PHYSICS|56 Videos

Similar Questions

Explore conceptually related problems

A body is projected at an angle 60^@ with horizontal with kinetic energy K. When the velocity makes an angle 30^@ with the horizontal, the kinetic energy of the body will be

If a body is projected with an angle theta to the horizontal, then

A ball is projected at an angle 60^(@) with the horizontal with speed 30 m/s. What will be the speed of the ball when it makes an angle 45^(@) with the horizontal ?

A particle is projected with velocity 50 m/s at an angle 60^(@) with the horizontal from the ground. The time after which its velocity will make an angle 45^(@) with the horizontal is

A body is projected at an angle of 30^@ with the horizontal and with a speed of 30 ms^-1 . What is the angle with the horizontal after 1.5 s ? (g = 10 ms^-2) .

A body is projected at an angle of 30^(@) with the horizontal and with a speed of 30ms^(-1) . What is the angle with the horizontal after 1.5 second? (g=10ms^(-2))

A particle is projected at 60(@) to the horizontal with a kinetic energy K . The kinetic energy at the highest point is

A body is projected at angle of 45^(@) to the horizontal . What is its velocity at the highest point ?

VIBRANT-REVIEW TEST-PART - II : PHYSICS
  1. At t=0 a projectile is fired from a point O (taken as origin) on the g...

    Text Solution

    |

  2. At t=0 a projectile is fired from a point O (taken as origin) on the g...

    Text Solution

    |

  3. A body is projected at an angle 60^(@) with the horizontal ground with...

    Text Solution

    |

  4. A rod can freely rotate in vertical plane about a hinge at its bottom....

    Text Solution

    |

  5. A bus is moving on a horizontal road with speed 5m//s. A & B are two p...

    Text Solution

    |

  6. A pendulum of mass m and length l [ as in figure] is attached to the t...

    Text Solution

    |

  7. A snail boat sails 2 km due East, 5 km 37^(@) South of East and finall...

    Text Solution

    |

  8. A block of mass M sits on an inclined plane and is connected via a mas...

    Text Solution

    |

  9. An electric sander has a continuous belt that rubs against a wooden su...

    Text Solution

    |

  10. A block of mass 1 kg start moving at t = 0 with speed 2 m//s on rough ...

    Text Solution

    |

  11. Assume that all the pulleys are massless and frictionless and strings ...

    Text Solution

    |

  12. A block of mass m(1) lies on top of fixed wedge as shown in figure 1 a...

    Text Solution

    |

  13. A pendulum is made by attaching a ball at the end of a string (the oth...

    Text Solution

    |

  14. A truck is travelling in a straight line on level ground, and is accel...

    Text Solution

    |

  15. Two equal masses m(1) = m(2) = 1 kg are kept on an inclined plane with...

    Text Solution

    |

  16. A car accelerates from rest at constant rate of 2 ms^(-2) for some tim...

    Text Solution

    |

  17. A sack of sand is lying on a light carpet on a horizontall surface. Wi...

    Text Solution

    |

  18. The surface below 2 kg is smooth but there is friction between the blo...

    Text Solution

    |

  19. A block of mass m is released on a smooth movable inclined plane of in...

    Text Solution

    |

  20. A wedge of height 'h' is released from rest with a light particle P pl...

    Text Solution

    |