Solutions: Option B Without the current in the river, the time taken for the swimmer to reach Q is 50s. This is based on the vertical constant speed of 1.2 m/s and veritcal (width of river) distance. With the current, which is constant 1.0 m/s, the swimmer will be drifted downstream. But the time taken to reach R from P is still 50s. So with the same 50s, you can calculate the horizontal distance based on the horizontal current speed of 1.0 m/s.

Solutions: Option B Key concepts: Both balls are free-falling, hence both will experience the same acceleration due to gravity 10 m/s2.

Solutions: Option B

A rock is thrown vertically upwards with a velocity of 29.4 m/s from the top of a building 78.4 m high. After how long will the rock reach the ground below? (Take g = 10 m/s2) Many are familiar with calculation when ball is released from rest. This question involves ball being thrown upwards, and then it falls to the ground. To solve this question, just treat the motion as 2 separate parts. 1) Ball going up to max height, 2) Ball going down (similar to ball released from rest from highest point)

The speed-time graph of Car A and Car B, along a straight road over 4 seconds is shown below. a) Calculate the acceleration of Car A and Car B over the 4 seconds.     Ans: a of Car A = 3 ms-2 and a of Car B =0.75 ms-2 Car A overtakes Car B at time t seconds. b) Derive two separate expressions for the velocities of Car A and Car B at the point when Car A overtakes Car B, in terms of t. Ans: Va = 3t    and    Vb = 3 + 0.75t c) Calculate the time t when Car A overtakes Car B.    Ans: t = 2.67s Solutions for (a) and (b) Solutions for (c)

Answer: Option A Refer to the video tutorial below for explanation.

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 F = 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 aa  = – 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) / 5v = 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.

Solutions: Option A

Answer: Option A

A U-tube with 30 cm of water is set up vertically and 12 cm of olive oil is poured carefully into the left-hand limb of the tube as shown below. It is observed that water will be pushed up in the other limb. The density of water and olive oil are 1000 kg m-3 and 920 kg m-3 respectively. Given A is at same level as the boundary between olive oil and water. a) What is the length of water column above point A? b) A further 6 cm of olive oil is added into the left-hand limb. How much further will the water level rise in the right limb? Solutions: View video tutorial for part (b) a) Poil = Pwater above A ρgh = ρgh920 x 10 x 12 = 1000 x 10 x h h = 11.04 cm b)         Pressure of water = pressure of olive oil 1000 x 10 x (2x + 11.04)/100  = 920 x 10 x 18/100 x = 2.76 cm