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# A cannon shell moving along a straight line bursts into two parts. Just after the burst one part moves with momentum $20\;{\text{N}} \cdot {\text{s}}$ making an angle $30^\circ$ with the original line of motion. The minimum momentum of the part of shell just after the burst is:A $0\;{\text{N}} \cdot {\text{s}}$B $5\;{\text{N}} \cdot {\text{s}}$C $10\;{\text{N}} \cdot {\text{s}}$D $17.32\;{\text{N}} \cdot {\text{s}}$

Last updated date: 18th Jun 2024
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Hint: The above problem can be solved by using the principle of conservation of the momentum. The momentum is the same as the inertia of the moving object. The momentum of the different objects before collision remains equal to the momentum of the objects after collision. The minimum momentum of the part shell just after the burst can be found by analyzing the motion of the cannon in horizontal and vertical direction.

Given: The momentum of the one part just after the burst is ${P_1} = 20\;{\text{N}} \cdot {\text{s}}$.
The angle of the one part with the original line of motion is $\alpha = 30^\circ$.
Let us assume that the minimum momentum of the part of the shell just after the burst is ${P_2}$.
The expression for the momentum of one part of the cannon shell in vertical direction is given as:
${P_{{y_1}}} = {P_1}\sin \alpha$
The expression for the momentum of other part of the cannon shell in vertical direction is given as:
${P_{{y_2}}} = {P_2}\sin \beta$
Apply the principle of conservation of the momentum in the vertical direction to find the minimum momentum of the part of the shell just after the burst.
${P_{{y_1}}} = {P_{{y_2}}}......\left( 1 \right)$
Substitute all the values in the expression (1) to find the minimum momentum of the part of the shell just after the burst.
${P_1}\sin \alpha = {P_2}\sin \beta$
Substitute $20\;{\text{N}} \cdot {\text{s}}$for ${P_1}$ and $30^\circ$ for $\alpha$ in the expression (2).
$\left( {20\;{\text{N}} \cdot {\text{s}}} \right)\left( {\sin 30^\circ } \right) = {P_2}\sin \beta$
${P_2}\sin \beta = 10\;{\text{N}} \cdot {\text{s}}......\left( 3 \right)$
The momentum of the other part becomes minimum only if the value of angle$\beta$ becomes maximum. The maximum value of sine angle becomes at $90^\circ$.
Substitute $90^\circ$ for $\beta$ in the expression (3) to find the minimum momentum of the part of the shell just after the burst.
${P_2}\left( {\sin 90^\circ } \right) = 10\;{\text{N}} \cdot {\text{s}}$
${P_2} = 10\;{\text{N}} \cdot {\text{s}}$

Thus, the minimum momentum of the part of the shell just after the burst is $10\;{\text{N}} \cdot {\text{s}}$and the option (C) is the correct answer.

Note: Be careful in applying the law of conservation of momentum. The law of conservation of momentum must be applied for all objects in a single direction at a time. If law applied for horizontal direction then it should be in horizontal direction for all the objects.