BPHET-141 Solved Assignment 2024

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BPHET-141 Solved Assignment 2024 Sample Solution
bphet-141-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

bphet-141-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

BPHET-141 Solved Assignment 2024 SS
PART A
  1. a) Light takes 5.2 years to travel from a distant planet to earth. If the astronaut travelled from earth to that planet at a speed of 0.8 c 0.8 c 0.8c0.8 \mathrm{c}0.8c, how long would it take according to the astronaut’s clock, to reach the planet?
Answer:
To find the time it takes for the astronaut to reach the planet according to the astronaut’s clock, we need to consider time dilation due to special relativity. The time measured by the astronaut, t ′ t ′ t^(‘)t’t′, is related to the time measured by an observer on Earth, t t ttt, by the following equation:
t ′ = t γ t ′ = t γ t^(‘)=(t)/( gamma)t’ = \frac{t}{\gamma}t′=tγ
where γ γ gamma\gammaγ is the Lorentz factor, given by:
γ = 1 1 − ( v c ) 2 γ = 1 1 − v c 2 gamma=(1)/(sqrt(1-((v)/(c))^(2)))\gamma = \frac{1}{\sqrt{1 – \left(\frac{v}{c}\right)^2}}γ=11−(vc)2
Here, v = 0.8 c v = 0.8 c v=0.8 cv = 0.8cv=0.8c is the speed of the astronaut, and c c ccc is the speed of light. The time it takes for light to travel from the planet to Earth is given as 5.2 years, so t = 5.2 years t = 5.2 years t=5.2″ years”t = 5.2 \text{ years}t=5.2 years.
First, let’s calculate the Lorentz factor:
γ = 1 1 − ( 0.8 ) 2 = 1 1 − 0.64 = 1 0.36 = 1 0.6 ≈ 1.667 γ = 1 1 − ( 0.8 ) 2 = 1 1 − 0.64 = 1 0.36 = 1 0.6 ≈ 1.667 gamma=(1)/(sqrt(1-(0.8)^(2)))=(1)/(sqrt(1-0.64))=(1)/(sqrt0.36)=(1)/(0.6)~~1.667\gamma = \frac{1}{\sqrt{1 – (0.8)^2}} = \frac{1}{\sqrt{1 – 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} \approx 1.667γ=11−(0.8)2=11−0.64=10.36=10.6≈1.667
Now, we can calculate the time it takes according to the astronaut’s clock:
t ′ = t γ = 5.2 years 1.667 ≈ 3.12 years t ′ = t γ = 5.2 years 1.667 ≈ 3.12 years t^(‘)=(t)/( gamma)=(5.2″ years”)/(1.667)~~3.12″ years”t’ = \frac{t}{\gamma} = \frac{5.2 \text{ years}}{1.667} \approx 3.12 \text{ years}t′=tγ=5.2 years1.667≈3.12 years
So, according to the astronaut’s clock, it would take approximately 3.12 years to reach the planet.
b) Calculate the linear momentum, total energy (in M e V M e V MeV\mathrm{MeV}MeV ) and kinetic energy (in M e V M e V MeV\mathrm{MeV}MeV ) of a particle travelling at 0.5 c 0.5 c 0.5c0.5 \mathrm{c}0.5c, given that its rest mass is 938 M e V 938 M e V 938MeV938 \mathrm{MeV}938MeV.
Answer:
To calculate the linear momentum, total energy, and kinetic energy of a particle traveling at 0.5 c 0.5 c 0.5 c0.5c0.5c, where c c ccc is the speed of light, and given that its rest mass is 938 M e V 938 M e V 938MeV938 \mathrm{MeV}938MeV (which corresponds to the rest mass of a proton), we can use the following relativistic formulas:
  1. Linear Momentum ( p p ppp):
    p = γ m 0 v p = γ m 0 v p=gammam_(0)vp = \gamma m_0 vp=γm0v
    where m 0 m 0 m_(0)m_0m0 is the rest mass, v v vvv is the velocity of the particle, and γ γ gamma\gammaγ is the Lorentz factor, given by:
    γ = 1 1 − v 2 c 2 γ = 1 1 − v 2 c 2 gamma=(1)/(sqrt(1-(v^(2))/(c^(2))))\gamma = \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}γ=11−v2c2
  2. Total Energy ( E E EEE):
    E = γ m 0 c 2 E = γ m 0 c 2 E=gammam_(0)c^(2)E = \gamma m_0 c^2E=γm0c2
  3. Kinetic Energy ( K K KKK):
    K = E − m 0 c 2 = ( γ − 1 ) m 0 c 2 K = E − m 0 c 2 = ( γ − 1 ) m 0 c 2 K=E-m_(0)c^(2)=(gamma-1)m_(0)c^(2)K = E – m_0 c^2 = (\gamma – 1) m_0 c^2K=E−m0c2=(γ−1)m0c2
Let’s calculate each of these quantities:
  1. Linear Momentum:
    γ = 1 1 − ( 0.5 c ) 2 c 2 = 1 1 − 0.25 = 1 0.75 = 1 0.866 ≈ 1.155 γ = 1 1 − ( 0.5 c ) 2 c 2 = 1 1 − 0.25 = 1 0.75 = 1 0.866 ≈ 1.155 gamma=(1)/(sqrt(1-((0.5 c)^(2))/(c^(2))))=(1)/(sqrt(1-0.25))=(1)/(sqrt0.75)=(1)/(0.866)~~1.155\gamma = \frac{1}{\sqrt{1 – \frac{(0.5c)^2}{c^2}}} = \frac{1}{\sqrt{1 – 0.25}} = \frac{1}{\sqrt{0.75}} = \frac{1}{0.866} \approx 1.155γ=11−(0.5c)2c2=11−0.25=10.75=10.866≈1.155
    p = ( 1.155 ) ( 938 M e V ) ( 0.5 ) = 542.09 M e V / c p = ( 1.155 ) ( 938 M e V ) ( 0.5 ) = 542.09 M e V / c p=(1.155)(938MeV)(0.5)=542.09MeV//cp = (1.155)(938 \mathrm{MeV})(0.5) = 542.09 \mathrm{MeV/c}p=(1.155)(938MeV)(0.5)=542.09MeV/c
  2. Total Energy:
    E = ( 1.155 ) ( 938 M e V ) = 1082.99 M e V E = ( 1.155 ) ( 938 M e V ) = 1082.99 M e V E=(1.155)(938MeV)=1082.99MeVE = (1.155)(938 \mathrm{MeV}) = 1082.99 \mathrm{MeV}E=(1.155)(938MeV)=1082.99MeV
  3. Kinetic Energy:
    K = ( 1.155 − 1 ) ( 938 M e V ) = 0.155 × 938 M e V = 145.39 M e V K = ( 1.155 − 1 ) ( 938 M e V ) = 0.155 × 938 M e V = 145.39 M e V K=(1.155-1)(938MeV)=0.155 xx938MeV=145.39MeVK = (1.155 – 1)(938 \mathrm{MeV}) = 0.155 \times 938 \mathrm{MeV} = 145.39 \mathrm{MeV}K=(1.155−1)(938MeV)=0.155×938MeV=145.39MeV
Therefore, the linear momentum of the particle is approximately 542.09 M e V / c 542.09 M e V / c 542.09MeV//c542.09 \mathrm{MeV/c}542.09MeV/c, the total energy is approximately 1082.99 M e V 1082.99 M e V 1082.99MeV1082.99 \mathrm{MeV}1082.99MeV, and the kinetic energy is approximately 145.39 M e V 145.39 M e V 145.39MeV145.39 \mathrm{MeV}145.39MeV.
c) Two spaceships of proper length L 0 L 0 L_(0)L_0L0 approach the earth from opposite direction at velocities ± 0.7 c ± 0.7 c +-0.7 c\pm 0.7 c±0.7c. What is the length of one of the spaceships with respect to the other?
Answer:
To find the length of one of the spaceships with respect to the other, we need to consider the relativistic effect of length contraction. When an object is moving relative to an observer, its length appears contracted along the direction of motion according to the formula:
L = L 0 1 − v 2 c 2 L = L 0 1 − v 2 c 2 L=L_(0)sqrt(1-(v^(2))/(c^(2)))L = L_0 \sqrt{1 – \frac{v^2}{c^2}}L=L01−v2c2
where L L LLL is the observed length, L 0 L 0 L_(0)L_0L0 is the proper length (the length of the object in its rest frame), v v vvv is the relative velocity between the observer and the object, and c c ccc is the speed of light.
In this case, the relative velocity between the two spaceships is the sum of their individual velocities, as they are moving in opposite directions. Therefore, v = 0.7 c + 0.7 c = 1.4 c v = 0.7 c + 0.7 c = 1.4 c v=0.7 c+0.7 c=1.4 cv = 0.7c + 0.7c = 1.4cv=0.7c+0.7c=1.4c. However, the maximum possible relative velocity cannot exceed the speed of light, so we need to use the relativistic velocity addition formula to find the correct relative velocity:
v rel = v 1 + v 2 1 + v 1 v 2 c 2 = 0.7 c + 0.7 c 1 + ( 0.7 c ) ( 0.7 c ) c 2 = 1.4 c 1 + 0.49 = 1.4 c 1.49 ≈ 0.9396 c v rel = v 1 + v 2 1 + v 1 v 2 c 2 = 0.7 c + 0.7 c 1 + ( 0.7 c ) ( 0.7 c ) c 2 = 1.4 c 1 + 0.49 = 1.4 c 1.49 ≈ 0.9396 c v_(“rel”)=(v_(1)+v_(2))/(1+(v_(1)v_(2))/(c^(2)))=(0.7 c+0.7 c)/(1+((0.7 c)(0.7 c))/(c^(2)))=(1.4 c)/(1+0.49)=(1.4 c)/(1.49)~~0.9396 cv_{\text{rel}} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} = \frac{0.7c + 0.7c}{1 + \frac{(0.7c)(0.7c)}{c^2}} = \frac{1.4c}{1 + 0.49} = \frac{1.4c}{1.49} \approx 0.9396cvrel=v1+v21+v1v2c2=0.7c+0.7c1+(0.7c)(0.7c)c2=1.4c1+0.49=1.4c1.49≈0.9396c
Now, we can use the length contraction formula to find the observed length of one spaceship with respect to the other:
L = L 0 1 − ( 0.9396 c ) 2 c 2 = L 0 1 − 0.8829 = L 0 0.1171 ≈ 0.3421 L 0 L = L 0 1 − ( 0.9396 c ) 2 c 2 = L 0 1 − 0.8829 = L 0 0.1171 ≈ 0.3421 L 0 L=L_(0)sqrt(1-((0.9396 c)^(2))/(c^(2)))=L_(0)sqrt(1-0.8829)=L_(0)sqrt0.1171~~0.3421L_(0)L = L_0 \sqrt{1 – \frac{(0.9396c)^2}{c^2}} = L_0 \sqrt{1 – 0.8829} = L_0 \sqrt{0.1171} \approx 0.3421 L_0L=L01−(0.9396c)2c2=L01−0.8829=L00.1171≈0.3421L0
So, the length of one of the spaceships with respect to the other is approximately 0.3421 L 0 0.3421 L 0 0.3421L_(0)0.3421 L_00.3421L0, meaning it appears to be about 34.21% of its proper length.
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