Question 35
Fig. 2.1 shows an
object M on a slope.
Fig. 2.1
M moves up the slope,
comes to rest at point Q and then moves back down the slope to point R.
M has a constant
acceleration of 3.0 m s-2 down
the slope at all times.
At time t
= 0, M is at point P and has a velocity of 3.6 m s-1 up the slope.
The total distance
from P to Q and then to R is 6.0 m.
(a)
Calculate, for the motion of M from P to Q,
(i)
the time taken, [2]
(ii)
the distance travelled. [1]
(b)
Show that the speed of M at R is 4.8 m s-1. [2]
(c)
On Fig. 2.2, draw the variation with time t
of the velocity v of M for the motion P
to Q to R.
Fig. 2.2
[3]
(d)
The mass of M is 450 g.
Calculate the
difference in the kinetic energy of M at P and at R. [2]
Reference: Past Exam Paper – November 2015 Paper 22 Q2
Solution:
(a)
(i)
v = u + at
{At Q, the
object comes to rest, so v = 0.
The motion
is up the slope while the acceleration is down the slope. So, the acceleration
is negative as it opposes the motion.}
0 = 3.6 –
3.0t
t (= 3.6 / 3.0) = 1.2 s
(ii)
{Consider
the motion from Q to P,
Average
speed = distance / time
(3.6+0)/2
= s / 1.2
s = 3.6 × 1.2 / 2}
(distance
to rest from P = (3.6 × 1.2) / 2 =) 2.2 (2.16) m
{Note that
the simple formula
speed =
distance / time
can only
be used when the speed is constant (there is no acceleration).
The
formula
Average speed
= total distance (being considered) / time
can be used
when there is a constant acceleration.}
(b)
{Total
distance from P to Q and then to R = 6.0 m
Distance
from Q to P = 2.16 m}
Distance
(from P to R) = 6.0 – 2.16 (= 3.84)
{From P
directly to R,}
v2 = u2 + 2as = 2 × 3.0 × 3.84 (= 23.04)
v2 = 3.62 + 2×1.68×3.0 (=
23.04)
v = 4.8 m s-1
(c)
straight
line from v = 3.6 m s–1 to v = 0 at t = 1.2 s
straight
line continues with the same gradient as v changes
sign
straight
line from v = 0 intercept to v = –4.8 m s-1
{The graph is
a straight line passing through (0, 3.6) {at P} to (1.2, 0) {at Q}. The
straight line continues up to v = - 4.8 m s-1 (at R).
It is a
straight line as the acceleration remains the same throughout the motion. SO,
the gradient is also constant.}
(d)
difference
in KE = ½ m(v2 – u2)
{Mass =
450 g = 0.45 kg}
difference
in KE = 0.5 × 0.45 (4.82 – 3.62) [= 5.184 –
2.916]
difference in KE = 2.3 (2.27) J
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