(B) At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component is $V \cos \theta$. Let the mass of

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(B) At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component is $V \cos \theta$. Let the mass of the shell be $2m$. According to the law of conservation of linear momentum in the horizontal direction:Initial momentum = Final momentum$(2m)(V \cos \theta) = m(0) + m(v_2)$Where $v_2$ is the velocity of the second piece.$2mV \cos \theta = mv_2$Therefore,$v_2 = 2V \cos \theta$.

Class 11 Physics Laws of Motion Conservation of Linear Momentum

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$A$ shell is fired from a cannon with a velocity $V$ at an angle $\theta$ with the horizontal direction. At the highest point in its path,it explodes into two pieces of equal masses. One of the pieces comes to rest. The speed of the other piece immediately after the explosion is

A
$3V \cos \theta$
B
$2V \cos \theta$
C
$\frac{3}{2} V \cos \theta$
D
$V \cos \theta$

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Conservation of Linear Momentum

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$A$ shell is fired from a cannon with a velocity $V$ at an angle $\theta$ with the horizontal direction. At the highest point in its path,it explodes into two pieces of equal masses. One of the pieces comes to rest. The speed of the other piece immediately after the explosion is

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$A$ bomb of mass $9\,kg$ explodes into $2$ pieces of mass $3\,kg$ and $6\,kg.$ The velocity of the $3\,kg$ mass is $1.6\,m/s.$ The kinetic energy of the $6\,kg$ mass is ............ $J.$

$A$ man of $50 \,kg$ is standing at one end of a boat of length $25 \,m$ and mass $200 \,kg$. If he starts running and when he reaches the other end, he has a velocity $2 \,ms^{-1}$ with respect to the boat. The final velocity of the boat is : (in $ms^{-1}$ )

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$A$ bomb initially at rest explodes by itself into three equal mass fragments. The velocities of two fragments are $(1\hat{i}+3\hat{j}) \ m/s$ and $(2\hat{i}-4\hat{j}) \ m/s$. The velocity of the third fragment (in $m/s$) is

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