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Question

A canon ball is fired with a velocity 200m/sec at an angle of 60° with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100m/sec, the second one falling vertically downwards with a velocity 100m/sec. The third fragment will be moving with a velocity,


  1. 100m/s in the horizontal direction

  2. 300m/s in the horizontal direction

  3. 300m/s in a direction making an angle of 60° with the horizontal

  4. 200m/s in a direction making an angle of 60° with the horizontal

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Solution

The correct option is C

300m/s in a direction making an angle of 60° with the horizontal


Step 1: Given data and drawing the diagram

The velocity of ball, u=200m/s

Angle with horizon =60°

The momentum of fragments =P1,P2,P3

Let the velocity of a third part after collision =V

Step 2: Find the momentum

The momentum of the ball (mass m) before explosion at the highest point,

=mvi^=mu(cos60°)i^=m×200×0.5i^=100mi^

[Since, at the maximum height, the initial velocity have only horizontal component as the vertical component becomes zero]

After the collision, momentum is,

=P1+P2+P3

=m3×100j^-m3×100j^+m3×Vi^

[According to the law of conservation of linear momentum]

Step 3: Find the velocity of third fragment

According to law of momentum,

Momentum before collision = momentum after collision

100mi^=m3×100j^-m3×100j^+m3×Vi^

V=300m/s

Hence, option (C) is correct.


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