Question 39: Centre of Mass and Rotational Motion

Question

A particle of mass m is projected with velocity v at an angle $$	heta$$ with the horizontal. At its highest point, it explodes into two pieces of equal mass, one of the piece continue to move on the original trajectory, then the velocity of second piece is

Accepted Answer

According to the law of conservation of linear momentum at the highest point,
$$mv\cos 	heta  = {m \over 2}(v\cos 	heta ) + {m \over 2}v'$$
$$ \Rightarrow mv\cos 	heta  - {m \over 2}v\cos 	heta  = {m \over 2}v'$$
$$ \Rightarrow v' = 2\left( {v\cos 	heta  - {{v\cos 	heta } \over 2}} ight) = v\cos 	heta $$

Answer

2 v cos$$	heta$$

Answer

v cos$$	heta$$

Answer

3 v cos$$	heta$$

Answer

$$\frac{v}{2}$$ cos$$	heta$$

Centre of Mass and Rotational Motion

A particle of mass m is projected with velocity v at an angle $\theta$ with the horizontal. At its highest point, it explodes into two pieces of equal mass, one of the piece continue to move on the original trajectory, then the velocity of second piece is

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