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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Tuesday, July 7, 2020

A radioactive substance contains a number of identical nuclei that emit β-particles. Which property of these nuclei remains unaltered by the emission?


Question 21
A radioactive substance contains a number of identical nuclei that emit β-particles.

Which property of these nuclei remains unaltered by the emission?
A charge
B neutron number
C nucleon number
D proton number





Reference: Past Exam Paper – June 2015 Paper 12 Q39





Solution:
Answer: C.

Beta particle is 0-1β.

The nucleon number of β-particles is zero. So, the nucleon number of the nucleus formed is the same as the parent nucleus.


Since the β-particle has a negative charge, the daughter nucleus will need to have a charge greater than the parent daughter so that the charge is conserved. Similarly, it will have one more proton than the parent nucleus.


The neutron number will also change as the proton number has changed.

Nucleon number = Neutron number + Proton number

Monday, June 29, 2020

Two stars A and B have their surfaces separated by a distance of 1.4 × 1012 m, as illustrated in Fig. 1.1


Question 9
(a) By reference to the definition of gravitational potential, explain why gravitational potential is a negative quantity. [2]


(b) Two stars A and B have their surfaces separated by a distance of 1.4 × 1012 m, as illustrated in Fig. 1.1.
 Fig. 1.1

Point P lies on the line joining the centres of the two stars. The distance x of point P from the surface of star A may be varied.
The variation with distance x of the gravitational potential φ at point P is shown in Fig. 1.2.

Fig. 1.2

A rock of mass 180 kg moves along the line joining the centres of the two stars, from star A towards star B.
(i) Use data from Fig. 1.2 to calculate the change in kinetic energy of the rock when it moves from the point where x = 0.1 × 1012 m to the point where x = 1.2 × 1012 m.
State whether this change is an increase or a decrease. [3]

(ii) At a point where x = 0.1 × 1012 m, the speed of the rock is v.
Determine the minimum speed v such that the rock reaches the point where x = 1.2 × 1012 m. [3]
[Total: 8]





Reference: Past Exam Paper – June 2016 Paper 41 & 43 Q1





Solution:
(a) The (gravitational) potential at infinity is defined as/is zero.
The (gravitational) force is attractive so work got out/is done as the object moves from infinity. (so the potential is negative)


(b)
(i)
{Gravitational potential is defined to be zero at infinity (and this is the maximum value).
The more negative the value, the smaller is the gravitational potential.

From the graph, it is observed that the gravitational potential is more negative at x = 1.2 × 1012 m (at B) compared to what it is at A. Thus, from A to B, there is an overall decrease in gravitational potential. This would appear as an increase in KE of the rock.}


{Change in potential energy: ΔE = mΔφ
The change in KE is equal to the change in PE.

From the graph,
At x = 1.2 × 1012 m, φ = – 14 × 108 J kg-1
At x = 0.1 × 1012 m, φ = – 10 × 108 J kg-1
Δφ = final φ – initial φ = – 14 × 108 – (– 10 × 108) = – 4 × 108 J kg-1

Below, only the magnitude is being considered as we have already determined that the KE is gained.
ΔE = mΔφ}


{Note that the equation ΔE = mΔφ = - GMm / r is valid for a body in the field of a single planet/starbody. When there are 2 stars (as in this case), we need to refer to the graph to know how the potential varies with position.}

ΔE = 180 × (14 – 10) × 108
ΔE = 7.2 × 1010 J
The change is an increase {in KE}.

(ii)
{The rock needs to reach the point where the field is zero. In the graph, this is where the gradient is zero.
Value of φ = – 4.4 × 108 J kg-1
The rock needs to reach only this point. The rock would be attracted by the field of the other planet afterwards.

ΔE = mΔφ}

Energy required = 180 × (10 – 4.4) × 108                  

{The KE is equal to this amount of energy.
½ mv2 = 180 × (10 – 4.4) × 108 }

½ × 180 × v2 = 180 × (10 – 4.4) × 108
Minimum speed v = 3.3 × 104 m s–1
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