1. The (net) force acting on an object is (directly) proportional to the rate of
change of momentum and takes place in the direction of the force. B2
[2]
2. According to Newton’s third law: When two objects interact, the force
acting on one of objects is equal but opposite to the force acting on the
other object. B1
The time t of ‘contact’ for the objects is the same and since Δp = Ft , B1
the gain in momentum for one object is equal to the loss of momentum for
the other object. B1
[3]
3. (i) u = 32 (m s–1) v = –2/3 × 32 = –21.33 (m s–1) t = 0.50 s C1
Δp = 800(–21.33 – 32) = –4.27 × 104 kg m s–1 A1
(ii) F = Δp / Δt = / F = 4.27 × 104 / 0.50 C1
F ≈ 8.5 × 104 (N) A1
Direction: Opposite to the initial velocity / away from the wall B1
(iii) N = pV / kT = (1.0 × 105 × 3.4 × 10–2) / (1.38 × 10–23 × 293) C1
N = 8.41 × 1023 C1
mass = 8.41 × 1023 / 6.02 × 1023) × 0.014
mass = 0.0196 (kg) ≈ 0.020 kg A1
[8]
4. (a) (i) to come to rest simultaneously, total mtm. = 0 or AW (1) 1
(but initial mtm. not zero)
(ii) initial mtm. = 3 m u – 2 m u = m u (1)
when closest, mtm. = (3m + 2m) v (1) 2
so 5m v = m u (and v = u / 5)
(b) (i) initial k.e. = final k.e. + (gain of) p.e. (1) 1
(ii) k.e. = ½ m v2 (1)
total k.e. = ½ × 3 m u2 + ½ × 2 m u2 (= 2.5 m u2) (1)
= 2.5 × 1.67 × 10–27 u2 (= 4.18 × 10–27 u2) (1) 3
allow m = 1.66 × 10–27 kg for full credit
(iii) gain of p.e. = initial k.e. – final k.e.
= 4.18 × 10–27 u2 – 4.18 × 10–27 (u/5)2 (2)
1.53 × 10–13 = 4.01 × 10–27 u2 (1) algebra
u = 6.18 × 106 m s–1 (1) 4
omits - 4.18 × 10–27 (u/5)2, gets u = 6. 06 × 106 m s–1: 1/2, 1, 1 = 3/4
[11]
5. (a) (i) Mass × velocity/mv with symbols defined 1
(ii) 0 = mAvA ± mBvB or mAvA = mBvB (1)
vA/vB = ± mB/mA (1) 2
max 1 mark for final expression without line 1
(b) (i) vA = (10/5 =) 2.0 (m s–1) and vB = (10/10 =) 1.0 (m s–1) 1
(ii) t1 = 3.0/2.0 = 1.5 (s) ecf b(i) 1
(iii) x = 2.1 – 1.0 × 1.5 = 0.6 (m) 1
(iv) v = vB + (5/50)vA = 1.0 + 0.2 (= 1.2 m s–1) 1
(v) t2 = t1 + 0.6/1.2 = 2.0 (s) 1
(vi) At collision the container (and fragments) stop (1)
By conservation of momentum, total momentum is still zero/AW (1) 2
(vii) straight lines from (0,0) to (1.5,0); (1.5,0) to (2.0,0.1); (x,0.1) for all x>2 3
[13]
6. (a) (i) 3.8 ± 0.3 (N s) 1
(ii) momentum (of the ball) accept impulse 1
(iii) mv = 3.8 or v = 3.8/0.16; = 23 (m s–1) ecf a 2
(iv) use F = ma giving 24 = 0.16a; a = 150 (m s–2) 2
(b) (i) exponential 1
(ii) e.g. h1/h2 = ek = 2.1(5); giving k = 0.74 to 0.76
or substitution from a line of table; gives 0.748, 0.757 or 0.746 2
(iii) 1.5 (m) 1
(iv) ∆k.e. = mg∆h; = 0.16 × 9.8 × 0.38 (= 0.60 J) 2
[12]
