A $6 \,kg$ bomb at rest explodes into three equal pieces $P, Q$ and $R$. If $P$ flies with speed $30 \,m / s$ and $Q$ with speed $40 \,m / s$ making an angle $90^{\circ}$ with the direction of $P$. The angle between the direction of motion of $P$ and $R$ is about

A body of mass $m_1$ moving with an unknown velocity of $v_1 \hat i$ undergoes a collinear collision with a body of mass $m_2$ moving with a velocity $v_2 \hat i$ . After collision $m_1$ and $m_2$ move with velocities of $v_3 \hat i$ and $v_4 \hat i$ respectively. If $m_2 = 0.5\, m_1$ and $v_3 = 0.5\, v_1$ then $v_1$ is:

The balls, having linear momenta $\vec{p}_1=\hat{p} \hat{i}$ and $\vec{p}_2=-p \hat{i}$, undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. The following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $\mathrm{p}, \mathrm{a}_1, \mathrm{a}_2, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{c}_1$ and $\mathrm{c}_2$.

$(A)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}$

$(B)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}$

$(C)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{p}}_2=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}$

$(D)$ $ \vec{p}_1^{\prime}=a_1 \hat{i}+b_1 \hat{j} $

$ \overrightarrow{\mathrm{p}}_2^{\prime}=a_2 \hat{\mathrm{i}}+b_1 \hat{\mathrm{j}}$

A man is standing on a balance and his weight is measured. If he takes a step in the left side, then weight

A jet engine works on the principle of

In an explosion a body breaks up into two pieces of unequal masses. In this