Unit 3: Work, Energy, and Power
Weightage : 15-25%
Two pucks moving on a frictionless air table are about to collide, as shown above. The \(1.5\,\mathrm{kg}\) puck is moving directly east at \(2.0\,\mathrm{m\,s^{-1}}\). The \(4.0\,\mathrm{kg}\) puck is moving directly north at \(1.0\,\mathrm{m\,s^{-1}}\).
What is the total kinetic energy of the two-puck system before the collision?
Correct Answer: \( \boxed{\mathrm{B}} \)
The total kinetic energy of the system is the sum of the kinetic energies of the two pucks.
Using
\(K=\dfrac{1}{2}mv^2\)
For the \(1.5\,\mathrm{kg}\) puck,
\(K_1=\dfrac{1}{2}(1.5)(2.0)^2=3.0\,\mathrm{J}\)
For the \(4.0\,\mathrm{kg}\) puck,
\(K_2=\dfrac{1}{2}(4.0)(1.0)^2=2.0\,\mathrm{J}\)
Therefore, the total kinetic energy is
\(K_{\mathrm{total}}=K_1+K_2=3.0+2.0=5.0\,\mathrm{J}\)
Thus, the correct answer is (B).
How does the work required to accelerate a particle from \(10\,\mathrm{m/s}\) to \(20\,\mathrm{m/s}\) compare to that required to accelerate it from \(20\,\mathrm{m/s}\) to \(30\,\mathrm{m/s}\)?
Correct Answer: \( \boxed{\mathrm{A}} \)
By the Work-Energy Theorem,
\(W=\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)\)
From \(10\,\mathrm{m/s}\) to \(20\,\mathrm{m/s}\),
\(W_1=\frac{1}{2}m(20^2-10^2)=\frac{1}{2}m(400-100)=150m\)
From \(20\,\mathrm{m/s}\) to \(30\,\mathrm{m/s}\),
\(W_2=\frac{1}{2}m(30^2-20^2)=\frac{1}{2}m(900-400)=250m\)
Since
\(250m>150m\),
more work is required to increase the speed from \(20\,\mathrm{m/s}\) to \(30\,\mathrm{m/s}\). Therefore, the work required to accelerate the particle from \(10\,\mathrm{m/s}\) to \(20\,\mathrm{m/s}\) is less.
If a particle moves in such a way that its position \(x\) is described as a function of time \(t\) by
\(x=t^{3/2}\),
then its kinetic energy is proportional to
Correct Answer: \( \boxed{\mathrm{C}} \)
The velocity is the time derivative of the position:
\(v=\dfrac{dx}{dt}=\dfrac{d}{dt}\left(t^{3/2}\right)=\dfrac{3}{2}t^{1/2}\)
Therefore,
\(v\propto t^{1/2}\)
The kinetic energy is
\(K=\dfrac{1}{2}mv^2\)
Since \(m\) is constant,
\(K\propto v^2\propto\left(t^{1/2}\right)^2=t\)
Hence, the kinetic energy is directly proportional to \(t\).
