
The total momentum of the system will be conserved before and after explosion.
Then, pi=pf
O=mvi^+mvj^+2mv
v=−2vi^−2vj^
∣v∣=2v
Total kinetic energy generated due to explosion is KE=2vmv2+2vmv2+2v2m(2v)2=mv2+2mv2=23mv2
NEET UG 2014 — Physics Mechanics
A body of mass (4m) is lying in x - y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (υ). The total kinetic energy generated due to explosion is:
Held on 30 Apr 2014 · Verified 9 Jul 2026.
mυ2
23mυ2
2mυ2
4mυ2
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A bullet of mass 10 g moving with 400 m/s penetrates a wall and comes to rest. The loss in kinetic energy is:
Consider a water tank shown in the figure. It has one wall at $x=L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0\left(\theta_0 \ll 1\right)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is:  $\left(\operatorname{take} \theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g\right.$ is the acceleration due to gravity)
A physical quantity $P$ is related to four observations $a, b, c$ and $d$ as follows : $P=a^3 b^2 / c \sqrt{d}$ The percentage errors of measurement in $a, b, c$ and $d$ are $1 \%, 3 \%, 2 \%$ and $4 \%$ respectively. The percentage error in the quantity $P$ is
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If $\mathrm{F}_{\mathrm{A}}$ and $\mathrm{F}_{\mathrm{B}}$ are the forces applied by the breaks on cars A and B , respectively, then the ratio $\mathrm{F}_{\mathrm{A}} / \mathrm{F}_{\mathrm{B}}$ is :
Work through every NEET UG Mechanics PYQ, year by year.