Conservation of Linear Momentum Questions in English

(C) The initial horizontal velocity of the object is $u_x = u \cos \theta$. Given $\sin \theta = 3/5$,then $\cos \theta = 4/5$. So,$u_x = 100 \times (4/5) = 80 \ m/s$.
At the highest point,the vertical velocity is $0$,and the horizontal velocity is $80 \ m/s$. The momentum of the system at the highest point is $P = (2m) \times 80 = 160m$.
After the explosion,the first piece $(m)$ comes to rest,so its velocity is $0$. Let the velocity of the second piece $(m)$ be $v$. By conservation of linear momentum: $160m = m(0) + m(v)$,which gives $v = 160 \ m/s$.
The time taken to reach the highest point is $t = (u \sin \theta) / g = (100 \times 3/5) / 10 = 6 \ s$. The horizontal distance covered to reach the highest point is $x_1 = u_x \times t = 80 \times 6 = 480 \ m$.
The second piece travels from the highest point to the ground in time $t = 6 \ s$ with a constant horizontal velocity of $160 \ m/s$. The additional horizontal distance covered is $x_2 = v \times t = 160 \times 6 = 960 \ m$.
The total distance from the point of projection is $x_1 + x_2 = 480 + 960 = 1440 \ m$.

(D) According to Newton's second law of motion,the rate of change of linear momentum is equal to the net external force acting on the system:
$F_{ext} = \frac{dp}{dt}$
If the total external force $F_{ext} = 0$,then $\frac{dp}{dt} = 0$,which implies that the linear momentum $p$ is constant (conserved).
In the case of a collision,if the time interval $\Delta t$ is extremely small (negligible) and the external force is finite,the impulse $J = F_{ext} \cdot \Delta t$ approaches zero.
Therefore,for a collision,the linear momentum is conserved if the external force is zero $OR$ if the external force is finite and the collision time is negligible.
Thus,both conditions $(1)$ and $(2)$ satisfy the requirement for the conservation of linear momentum.

(D) Let the mass of the first piece be $m_1 = 4 \,kg$ and its velocity be $v_1$.
Let the mass of the second piece be $m_2 = 12 \,kg$ and its velocity be $v_2 = 4 \,ms^{-1}$.
According to the law of conservation of linear momentum, the initial momentum of the bomb is zero (assuming it was at rest).
Therefore, $m_1 v_1 + m_2 v_2 = 0$.
Taking magnitudes, $m_1 v_1 = m_2 v_2$.
$4 \times v_1 = 12 \times 4$.
$v_1 = 12 \,ms^{-1}$.
The kinetic energy of the $4 \,kg$ piece is $KE_1 = \frac{1}{2} m_1 v_1^2$.
$KE_1 = \frac{1}{2} \times 4 \times (12)^2 = 2 \times 144 = 288 \,J$.

(D) Let the mass of the shell be $M$ and its velocity at the highest point be $v_h = u \cos \theta$. The momentum at the highest point is $P = M u \cos \theta$.
After breaking into two equal parts of mass $M/2$,one part retraces its path,meaning its velocity is $-u \cos \theta$. Let the velocity of the second part be $v_2$.
By conservation of linear momentum:
$M u \cos \theta = \frac{M}{2} (-u \cos \theta) + \frac{M}{2} v_2$
$u \cos \theta = -\frac{1}{2} u \cos \theta + \frac{1}{2} v_2$
$\frac{3}{2} u \cos \theta = \frac{1}{2} v_2 \implies v_2 = 3 u \cos \theta$.
The kinetic energy of the first part is $E_1 = \frac{1}{2} (M/2) (u \cos \theta)^2 = \frac{1}{4} M u^2 \cos^2 \theta$.
The kinetic energy of the second part is $E_2 = \frac{1}{2} (M/2) (3 u \cos \theta)^2 = \frac{1}{4} M (9 u^2 \cos^2 \theta) = \frac{9}{4} M u^2 \cos^2 \theta$.
Comparing the two,we get $E_2 = 9 E_1$.

(C) Let the velocity of the third fragment be $v'$.
According to the law of conservation of linear momentum,since no external force acts on the system,the total initial momentum is equal to the total final momentum.
Initial momentum $P_i = 5 M \times 0 = 0$.
Let the velocity of the first fragment $(M)$ be $\vec{v}_1 = 2v \hat{i}$ and the velocity of the second fragment $(2M)$ be $\vec{v}_2 = -v \hat{i}$ (since they move in opposite directions).
Final momentum $P_f = M(2v \hat{i}) + 2M(-v \hat{i}) + 2M(\vec{v}') = 0$.
$2Mv \hat{i} - 2Mv \hat{i} + 2M(\vec{v}') = 0$.
$0 + 2M(\vec{v}') = 0$.
Therefore,$\vec{v}' = 0$.
The third fragment will be at rest.

(B) The total mass of the body is $M = 14 \ kg$. The masses of the three fragments are in the ratio $2:2:3$. Let the masses be $m_1 = 4 \ kg$,$m_2 = 4 \ kg$,and $m_3 = 6 \ kg$ (since $2x + 2x + 3x = 7x = 14 \ kg$,so $x = 2 \ kg$).
Initially,the body is at rest,so the initial momentum is zero. By the law of conservation of linear momentum,the final total momentum must also be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.
Let the two equal fragments move along the negative $x$-axis and negative $y$-axis respectively: $\vec{v}_1 = -18 \hat{i} \ m/s$ and $\vec{v}_2 = -18 \hat{j} \ m/s$.
The momentum of the first two fragments is $\vec{p}_1 + \vec{p}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2 = 4(-18 \hat{i}) + 4(-18 \hat{j}) = -72 \hat{i} - 72 \hat{j} \ kg \cdot m/s$.
Since $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$,we have $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = 72 \hat{i} + 72 \hat{j} \ kg \cdot m/s$.
The velocity of the third (heavier) fragment is $\vec{v}_3 = \frac{\vec{p}_3}{m_3} = \frac{72 \hat{i} + 72 \hat{j}}{6} = 12 \hat{i} + 12 \hat{j} \ m/s$.
The magnitude of the velocity is $|\vec{v}_3| = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \ m/s$.