Below are the questions.
2003PPq1q7 (0 min to 2.42 min) Ans: Option A
2023PPp1q4 (2.47 min 3.54 min) Ans: Option C
2018PPp1q2 (3.45 min to 6.22 min) Ans: Option D
2011PPp1q6, 2016PPp1q6 (6.22 min) Ans: Option A
A force of 3.0 N and a force of 4.0 N act on an object at the same time. Which of the following forces cannot be the resultant force on the object?
A) 0.5 N B) 3.5 N C) 4.0 N D) 5.0 N
Solutions: A) 0.5 N
To achieve maximum resultant force from 3.0 N and 4.0 N, both forces have to be parallel and in the same direction. Hence the maximum resultant force will be 3 N + 4 N = 7 N
To achieve minimum resultant force from 3.0 N and 4.0 N, both forces have to be parallel and in the opposite direction. Hence the minimum resultant force will be 4 N – 3 N = 1 N
Hence any magnitude of resultant force between 1 N and 7 N is possible as the 2 forces can be positioned at an angle to each other.
Therefore, only option (A) cannot be the resultant force on the object as it is less than the minimum resultant force of 1 N.
A ball is thrown vertically upwards from the hand and lands back onto the hand. It is important to note that once the ball leaves the hand, the resultant force acting on the ball is only its weight! And it is acting downwards throughout the motion.
Velocity-Time Graph
Displacement-Time Graph
Key points to note when sketching the v-t graph:
1) Fixing the direction up to be positive (can be down as positive if you want)
2) At t = 0s, the initial velocity of the ball is maximum.
3) As it goes up vertically, due to the weight acting downwards, the ball decreases in speed (decelerates)
4) At the highest point, the ball is momentarily at rest (v =0m/s)
5) On the way down, due to the weight of the ball acting downward, the ball accelerates downwards.
A 10 kg box is initially at rest on a rough horizontal surface and a horizontal force is then applied to it. The force varies as shown in the diagram.
a) If the box moves with an acceleration of 2m/s2 in the first 15 s, what is the frictional force between the surfaces?
Resultant force on box = ma
Applied force – frictional force = ma
40 – frictional force = 10 x 2
frictional force = 20 N
b) How does the velocity change during the next 5 s?
Next 5 s after time 15 s, there is no applied force. Hence the resultant force acting on box isonly frictional force 20N (in direction opposite to motion of box).
F = ma
-20 = 10 x a
a = – 2 m/s2
The velocity of box at time 15s is required.
Hence when t = 0s, box accelerates at 2m/s2 from rest,
a = (v – u) / t
2 = (v – 0) /15
v = 30 m/s
(Hence the final velocity of the box at time 15s is 30 m/s, just before applied force of 40 N is released)
Find velocity of box at end of next 5s (time = 20s):
a = (v – u) /t
– 2 = (v – 30) / 5
v = 20 m/s.
Therefore, velocity decreases at a constant rate of -2 m/s2 from 30 m/s to 20 m/s in next 5s.
c) Plot a velocity-time graph and calculate the distance travelled by the box in the 30 s.
A trolley is acted on by three forces, P, Q and R in magnitude and direction as shown in the figure below. A single force F could balance all three forces. What is the magnitude of F?
Solutions:There are 3 different forces, P, Q and R acting on the trolley.Consider the forces P and R only, as they are acting perpendicular to each other, the resultant force falls exactly on Q. (using the vector diagram method/parallelogram method).The resultant of P and R is Q, plus the existing Q, will give a final resultant force of 2Q acting on the trolley.
Evan's Space 