IBDP Physics 10.2 – Fields at work: IB Style Question Bank HL Paper 1

Question

A planet has radius R. The escape speed from the surface of the planet is v. At what distance from the surface of the planet is the orbital speed 0.5v?

A 0.5R

B R

C 2R

D4R

Answer/Explanation

Ans: B

ESCAPE SPEED (Ve)

It is the minimum speed with which a body should be projected from the surface of a planet so as to reach at infinity i.e., beyond the gravitational field of the planet.
If a body of mass m is projected with speed v from the surface of a planet of mass M and radius R, then

ORBITAL VELOCITY (V0 )

Let a satellite of mass m revolve around the planet in circular orbit of radius r with speed v0. The gravitational pull between satellite and planet provides the necessary centripetal force.
Orbital velocity,
\(v_o=\sqrt{\frac{GM}{r}}\)
Now given that
\(v_o=0.5v_e\)
hence
\(\sqrt{\frac{GM}{r}} = 0.5 \sqrt{\frac{2GM}{R}} =\frac{1}{2} \sqrt{\frac{2GM}{R}}\)
or
\(\frac{GM}{r}=\frac{1}{4}\times \frac{2GM}{R}\)
\(r =2R\)
hence from surface of planet distance \(= 2R-R =R\)

Question

A particle with charge – 2.5 × {10}^{-6} C moves from point X to point Y due to a uniform electrostatic field. The diagram shows some equipotential lines of the field.

What is correct about the motion of the particle from X to Y and the magnitude of the work done by the field on the particle?

Ans:

Motion of the particle from X to Y

Magnitude of the work done by the

field on the particle

A

uniform linear

0 J

B

uniform linear

1 J

C

 uniformly accelerated

0 J

D

uniformly accelerated

1 J

Answer/Explanation

Ans: D

Force \(F_q = qE\) is from point X towards Y as q is negative. Since E is uniform , hence F is uniform and hence particle accelerate uniformly.

Now

Work done  \(W= q \Delta V \)

\(W= q\Delta V = -2.5 \times 10 ^{-6}(700-300)\times 10^3 =-1000\times 10^{-6}=-1 J \)
Magnitude of work done by field \(= |- W|= 1J \)

Question

A satellite orbiting a planet moves from orbit X to orbit Y.

                                                       

What is the change in the kinetic energy and the change in the gravitational potential energy as a result?

Answer/Explanation

Markscheme

C

\(K.E = \frac{1}{2}mv_o^2 where v_o^2 = \frac{GM}{r}\)
\(\therefore K.E =\frac{GMm}{2r}\)  hence as  \(r \;\uparrow K.E \; \downarrow\)
Potential Energy is given by  \(\frac{-GMm}{r}\)
Hence as  \(r \; \uparrow \; P.E  \;\uparrow \)(since it become less negative)

Question

An electron of mass me orbits an alpha particle of mass mα in a circular orbit of radius r. Which expression gives the speed of the electron?

A. \(\sqrt {\frac{{2k{e^2}}}{{{m_e}r}}} \)

B. \(\sqrt {\frac{{2k{e^2}}}{{{m_a}r}}} \)

C. \(\sqrt {\frac{{4k{e^2}}}{{{m_e}r}}} \)

D. \(\sqrt {\frac{{4k{e^2}}}{{{m_a}r}}} \)

Answer/Explanation

Markscheme

A

Centripetal force = Electrostatic attraction force (since Electron is in circular orbit)

\(\frac{m_ev^2}{r}=\frac{k (e)(2e)}{r^2}\)
Alpha particle is  \({He}^{++}\)
Hence
\(v^2 =\frac{k (e)(2e)}{m_er}\)
or
\(v=\sqrt{\frac{k (e)(2e)}{m_er}}=\sqrt{\frac{2k e^2}{m_er}} \)

Question

A spacecraft moves towards the Earth under the influence of the gravitational field of the Earth.

The three quantities that depend on the distance r of the spacecraft from the centre of the Earth are the

I.   gravitational potential energy of the spacecraft
II   gravitational field strength acting on the spacecraft
III. gravitational force acting on the spacecraft.

Which of the quantities are proportional to \(\frac{1}{{{r^2}}}\)?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Answer/Explanation

Markscheme

C

 gravitational potential energy

\(U_g =\frac{-GM_em_s}{r}\)
gravitational field strength
\(E_g =\frac{GM_e}{r^2}\)
gravitational force
\(F_g=\frac{GM_em_s}{r^2}\)

Here We can See gravitational field strength and gravitational force depends on \(r\) as \(\frac{1}{{{r^2}}}\)

Question

Four uniform planets have masses and radii as shown. Which planet has the smallest escape speed?

Answer/Explanation

Markscheme

C

Escape velocity is given by

\(v_e=\sqrt{\frac{2GM_p}{R_p}}\)
For smallest  \(v_e , M_p \) should be small and  \(R_p \)  should be big
\(\frac{M_p}{R_p}\)
for \(A = 1\) , For \(B =2\) , for \(C = \frac{1}{2} \) and for  \(D =1\)
Hence C will have least escape velocity

Question

The graph shows the variation of the gravitational potential V with distance r from the centre of a uniform spherical planet. The radius of the planet is R. The shaded area is S.

What is the work done by the gravitational force as a point mass m is moved from the surface of the planet to a distance 6R from the centre?

A. m (V2 – V1 )

B. m (V1 – V2 )

C. mS

D. S

Answer/Explanation

Markscheme

B

Gravitational potential is given by 

W= -m\Delta V as gravitational field is conservative field and hence work done is independent of path. negative sign because force and dispalcement are in opposite direction
W=- m(V_2-V_1)=m(V_1 -V_2)