MC

Q1. This is like the one we did in class from the momentum section.

Area under graph is impulse so Ft = 0.5 x 10 x 40 = 200 Ns

That is the change in momentum - initial momentum being zero we know that final momentum = 200 Ns

mv = 200 Ns

m = 0.25 kg therefore v = 200/0.25 = 800 m/s

Q2. Conservation of momentum

Before - both at rest

After - they go in opposite directions

daughter nucleus has mas of M-m

(M-m) v₁ = mv₂

E = mv₂²/2

V₂ = Root(2E/m)

so v₁ = m/(M-m) x Root(2E/m) - square m and put it in the root area of the equation - then cancel

v₁ = Root(2Em)/(M-m) - choice A

Q3.

Momentum is conserved so unchnged in first column - both moving later so KE increases. Choice A

 

Q4. distance travelled = 2πr = 2π x 1.5 x 10⁸ km = 2π x 1.5 x 10¹¹ m

time taken = 365.25 x 24 x 60² seconds

speed = 2π x 1.5 x 10¹¹/365.25 x 24 x 60² = 3.0 x 10⁴ m/s

 

Q5. constant speed therefore constant KE

 

Q6. They have used this one before... see the circuilat motion questions.

The tension in the string is due to the centripetal force of the mass = mv2/r

This is balanced by the weight of the mass Mg

∴Mg = mv2/r

so M = mv2/rg

Q7. The weight acts all of the time - tension in the spring due to oscillation is zero at the centre of oscillation so Force must minimum at that point

 

Q8. m = 10 kg so w = 98.1 N extension = 0.25 m

k = force/extension = 9.81/0.25 m

 

T = 2π ROOT(m/k) = 2π ROOT(10 x 0.25/98.1) =1.0s (choice A)

 

Q9. This has been used before too... see SHM questions.

For the pendulum g is inversely proportional to T².

Therefore if g halved the period would increase (by a factor of root 2).

The spring system's period does not depend on g. It depends on the mass (which is the same wherever the body is) and the spring constant. The spring constant depends on the internal arrangment of the spring's particles. That is unaltered by position in a graviational field.

 

Q10. Resonance produces increased amplitude at natural frequency therefore it must be B or D. Lightest damping means you will get quite a big response at the natural frequency. Therefore it is B.

 

Q11. D is wrong - it is r²!

 

Q12. weight is due to gravitational force

W depends directly on the mass of the planet and inversely on the square of its radius

W = constant x M/R²

constant = WR²/M

W₂ = WR²/M x 2M/(2R)² = W/2

 

Q13. g = GM/r²

density is the same so mass will increase as volume does:

M₂ = M₁ x 100³

r₂ = r₁ x 100

g₂/g₁ = M₁r₂²/M₂r₁² = 100² x M₁ x (r₁ x 100)²/( M₁ x 100³ x r₁² )

g₂/g₁ = 100² x 100²/100³ = 100

g₂ = 100 g₁ = 800 N/kg

Q14. Think of the planets - inner ones whiz round - outer ones move slowly. KE increases as it is closer to the gravitational centre. - A is the choice.

 

Q15. F depends on inverse of r².

triple r and F drops to a ninth.

 

Q16. P is minimum approach - max potential min kinetic - choice C

 

Q17. E = 1/2 CV2

if energy is constant

2E = C₁V₁² = C₂V₂²

V₁²/V₂² = C₂/C₁ = 10/1000 = 1/100 - choice B

 

Q18.

time constant = CR = 10 x 10^-3 x 500 = 5.0 s

t = RC = 5.0 seconds

V = V₀ e ^-1 = 6e ^-1 = 2.2 V (choice A or D)

1000 readings in 10 seconds so 500 in 5s - choice D

 

Q19.

V = V₀ e ^-92/CR

1.5 = 6.0 e ^-92/CR

ln (0.25) = -92/(220 x 10^-6 x R)

R = -92/(220 x 10^-6 x ln (0.25) ) = -92/(-3.05 x 10^-4) = 3.0 x 10⁵ Ω

 

Q20.

F = BIl

F/BI = constant

F₂ = F₁B₂I₂/B₁I₁ = F x 2 x 0.25 = F/2 (Choice B)

 

Q21.

Force on a charge FLHR

- middle finger in direction of conventional current or positive charge (to the right of page)

- Thumb down

- Forefinger (field) is then out of page - solution A

 

Q22. Induced emf will oppose the motion of the magnet - slow it down. The one with the lowest resistance will produce the biggest current and therefore slow it most.

Copper has lowest resistance therefore will emerge last - rubber won't produce an emf to oppose motion as it isn;t a metal Order is RQP - choice D

 

Q23. ε = N ΔΦ/Δt

so the shape would be that of the gradient of the curve - the curve gradient is positive and increasing so it must be D

 

Q24. Induced current in second could depends on how quickly the magnetic field changes around coil X. The most rapid change will be when it was switched off - choice C.

 

Q25.

IpVp/IsVs = 0.15

(350 x 10^-3 x 230)/(50 x 10^-3 x 0.15) = Vs = 4.93 V - choice A