Question
A particle of mass 0.4 kg is released from rest at a height of 1.8 m above the surface of the water in a
tank. There is no instantaneous change of speed when the particle enters the water. The water exerts an upward force of 5.6 N on the particle when it is in the water.
(i) Find the velocity of the particle at the instant when it reaches the surface of the water.
(ii) Find the time that it takes from the instant when the particle enters the water until it comes to instantaneous rest in the water. You may assume that the tank is deep enough so that the particle does not reach the bottom of the tank.
(iii) Sketch a velocity-time graph for the motion of the particle from the instant at which it is released
until it comes to instantaneous rest in the water.
Answer/Explanation
(i) \(0.4\times 1.8=\frac{1}{2}\times 0.4\times \nu ^{2}\)
\(\nu =6ms^{-1}\)
Alternative method for question (i)
\(\nu ^{2}= 0^{2}+2\times g\times \times 1.8
v=6ms^{-1}\)
(ii) 0.4g – 5.6 = 0.4a
\(a=-4ms^{-2}\)
0=6-4t t=1.5 s
(iii) Straight line starting at (0,0) with positive gradient Second straight line starting at end of the first line with negative
gradient and ending with
v = 0
All correct, start at (0, 0) with max velocity v= 6 at t = 0.6 i.e. (0.6, 6) and finishing at (2.1, 0)
Question
An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments.
The elevator accelerates upwards from rest to a speed of \(2ms^{-1}\) over a period of 1.5s and then travels at this speed for before decelerating to rest over a period of 1 s.
The elevator then remains at rest for 6 s. before accelerating to a speed of \( Vms^{-1}\) downwards over a period of 2s. The elevator travels at this speed for a period of 5s, before decelerating to rest over a period of 1.5s.
(a) Find the acceleration of the elevator during the first 1.5s.
(b) Given that the elevator starts and finishes its journey on the ground floor, find V.
(c) The combined weight of the elevator and passengers on its upward journey is 1500kg. Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating.
Answer/Explanation
Ans:
- Acceleration \(=\frac{4}{3}ms^{-2}\)
- \(\frac{1}{2}(7+4.5)\times 2 = \frac{1}{2}(8.5 + 5) \times V\)
V=1.7[0] (3sf) - Acceleration \(=-2ms^{-2}\)
\(T-1500g=1500 \times (-2)\)
T=12000N
Question
The velocity of a particle moving in a straight line is v \(ms^{-1}\)at time t seconds after leaving a fixed
point O. The diagram shows a velocity-time graph which models the motion of the particle from t = 0to t = 16. The graph consists of five straight line segments. The acceleration of the particle from t = 0 to t = 3 is 3 \(ms^{-2}\)
The velocity of the particle at t = 5 is 7 m s^{-1} and it comes to instantaneous rest at t = 8. The particle then comes to rest again at t = 16. The minimum velocity of the particle is V m s^{-1}.
(i) Find the distance travelled by the particle in the first 8 s of its motion.
(ii) Given that when the particle comes to rest at t = 16 its displacement from O is 32 m, find the value of V.
Answer/Explanation
5(i) Velocity at t = 3 is 3 × 3 = 9
[1⁄2 × 3 × 9 + 1⁄2 (9 + 7) × 2 + 1⁄2 × 3 × 7]
Distance = area under graph.
Distance = 40 m
5(ii) [32 = 40 + area of triangle] Area of triangle or displacement/distance =(–)8[Distance = 1⁄2 × 8 ×V = (–)8] V = –2
Question
A particle P moves in a straight line starting from a point O and comes to rest 35 s later. At time ts after leaving O, the velocity v \(ms^{-1}\) of P is given by
where a and b are constants such that a > 0 and b < 0.
(i) Show that the values of a and b are 49 and −0.04 respectively.
(ii) Sketch the velocity-time graph.
(iii) Find the total distance travelled by P during the 35 s.
Answer/Explanation
Question
The diagram shows a velocity-time graph which models the motion of a cyclist. The graph consists
of five straight line segments. The cyclist accelerates from rest to a speed of 5\( ms^{-1}\) over a period of 10 s, and then travels at this speed for a further 20 s. The cyclist then descends a hill, accelerating to speed V\( m s^{-1}\)over a period of 10 s. This speed is maintained for a further 30 s. The cyclist thendecelerates to rest over a period of 20 s.
(i) Find the acceleration of the cyclist during the first 10 seconds.
(ii) Show that the total distance travelled by the cyclist in the 90 seconds of motion may be expressed
as(45V + 150) m. Hence find V, given that the total distance travelled by the cyclist is 465 m. (iii) The combined mass of the cyclist and the bicycle is 80 kg. The cyclist experiences a constant resistance to motion of 20 N. Use an energy method to find the vertical distance which the cyclist descends during the downhill section from t = 30 to t = 40, assuming that the cyclist does no work during this time.
Answer/Explanation
(i) a =\( 0.5ms^{-2}\)
(ii) [Distance
= 25 + 100 + 5(5 + V) + 30V + 10V]
150 + 45V
150 + 45V = 465 → V = \(7ms^{-1}\)
(iii)\( 1/2\times 80\times 7^{2}-1/2\times 80\times 5^{2}[=960]\)
20 × (5 + 7)/2 × 10 [=1200]
[80gh = 960 + 1200]
h = 2.7m