Fig. 2.1 shows an object M on a slope. M moves up the slope, comes to rest at point Q and then moves back down the slope to point R.

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