IMO Practice Test: Work, Energy and Power | Class 11 Physics

Question 1 of 12

A body of mass $m$ moves so that its position is $x=3t^2$ (x in metres, t in seconds). The work done by the net force in the first 2 s is:

$18m\ \text{J}$

$36m\ \text{J}$

$72m\ \text{J}$

$9m\ \text{J}$

Explanation: $v=\frac{dx}{dt}=6t$, so at $t=2$, $v=12\ \text{m/s}$, $u=0$. $W=\Delta KE=\frac{1}{2}m(12)^2=72m\ \text{J}$.

Question 2 of 12

A force $F=kx$ acts on a particle. The work done in moving it from $x=0$ to $x=a$ is:

$ka^2$

$\frac{1}{2}ka^2$

$2ka^2$

$ka$

Explanation: $W=\int_0^a kx\,dx=\frac{1}{2}ka^2$.

Question 3 of 12

Two bodies of equal mass moving with equal and opposite velocities $v$ undergo a perfectly inelastic head-on collision. The KE lost is:

zero

$\frac{1}{2}mv^2$

$mv^2$

$2mv^2$

Explanation: Total momentum is zero, so they stop; initial KE $=2\times\frac{1}{2}mv^2=mv^2$ is fully lost.

Question 4 of 12

A spring of constant $k$ is cut into two equal halves. The spring constant of each half is:

$\frac{k}{2}$

$k$

$2k$

$4k$

Explanation: Spring constant is inversely proportional to length; halving the length doubles $k$ to $2k$.

Question 5 of 12

A car of mass $m$ moving at speed $v$ is stopped in distance $d$ by braking. To stop the same car at speed $2v$ with the same retarding force, the distance needed is:

$2d$

$3d$

$4d$

$d$

Explanation: $Fd=\frac{1}{2}mv^2$, so $d\propto v^2$; doubling $v$ makes the distance $4d$.

Question 6 of 12

The power of a body moving with velocity $v$ under a constant force $F$ along its direction is:

$\frac{F}{v}$

$Fv$

$Fv^2$

$\frac{1}{2}Fv$

Explanation: $P=\vec{F}\cdot\vec{v}=Fv$ when force is along the velocity.

Question 7 of 12

A 1 kg ball moving at 4 m/s collides elastically and head-on with a 3 kg ball at rest. The velocity of the 1 kg ball after collision is:

$-2\ \text{m/s}$

$+2\ \text{m/s}$

$-4\ \text{m/s}$

$+1\ \text{m/s}$

Explanation: $v_1=\frac{(m_1-m_2)u_1}{m_1+m_2}=\frac{(1-3)\times 4}{1+3}=\frac{-8}{4}=-2\ \text{m/s}$ (it bounces back).

Question 8 of 12

A particle moves under a force such that its KE is proportional to time, $KE\propto t$. Its speed is proportional to:

$t$

$t^2$

$\sqrt{t}$

constant

Explanation: $KE=\frac{1}{2}mv^2\propto t$ gives $v^2\propto t$, so $v\propto\sqrt{t}$.

Question 9 of 12

A ball is dropped from height $h$ and the coefficient of restitution with the floor is $e$. The height after the first bounce is:

$eh$

$e^2h$

$\frac{h}{e}$

$\sqrt{e}\,h$

Explanation: Rebound speed is $e$ times the approach speed, so rebound height $=e^2h$ (height $\propto v^2$).

Question 10 of 12

An engine pumps water at a rate $\dot{m}$ (kg/s) giving it speed $v$. The power delivered to the water is:

$\dot{m}v$

$\frac{1}{2}\dot{m}v^2$

$\dot{m}v^2$

$\frac{1}{2}\dot{m}v$

Explanation: KE given per second $=\frac{1}{2}(\dot{m})v^2$, which is the power delivered.

Question 11 of 12

A block of mass 2 kg is pushed against a spring ($k=800\ \text{N/m}$) compressing it 0.1 m, then released on a frictionless floor. The block's speed when it leaves the spring is:

$1\ \text{m/s}$

$2\ \text{m/s}$

$4\ \text{m/s}$

$0.5\ \text{m/s}$

Explanation: $\frac{1}{2}kx^2=\frac{1}{2}mv^2$, so $v=x\sqrt{\frac{k}{m}}=0.1\sqrt{\frac{800}{2}}=0.1\times 20=2\ \text{m/s}$.

Question 12 of 12

In a 1D elastic collision, the relative velocity of separation compared with the relative velocity of approach is:

greater

smaller

equal in magnitude

zero

Explanation: For a perfectly elastic collision, $u_1-u_2=v_2-v_1$ — the relative velocities of approach and separation are equal in magnitude.

IMO Practice Test — Work, Energy and Power