Q1 B
Q2 B
Q3 D
Q4 B
Q5 D
Q6 D
M7.
(a) kinetic energy is not conserved (1)
(or velocity of
approach equals velocity of separation)
1
(b)
(i) (use of p = mv gives) p = 4.5 ×
10–2 × 60 (1)
8. (i) Force
is proportional to the rate of change of momentum
(QWC This mark can only be scored if momentum is spelled
correctly)
Allow
“equal” instead of proportional, allow “change in momentum over time” (WTTE)
Do not allow F = ma or in words
B1
(ii) When
one body exerts a force on another the other body exerts an equal
(in magnitude) and opposite (in direction) force on the first body
(WTTE)
Must
refer to two bodies.
Do not allow a bare “Action and reaction are equal and opposite”.
B1
[2]
9. kinetic
energy is the energy a body possesses by virtue of its speed (1)
s an energy it is a measure of force × distance (1)
the rate of change of momentum defines force (1)
momentum is therefore a measure of force × time (1) 4
Other possible answers will score a maximum
of 3 unless the
force × distance relationship is given for kinetic energy and the
force × time relationship is given for momentum
momentum is always conserved in
a collision (in the absence of
external forces) (1)
but kinetic energy may be lost – with qualification of what happens (1)
kinetic energy is proportional to v2 but momentum is proportional to v (1)
kinetic energy is a scalar; momentum is a vector (1)
[4]
10. (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]
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