A constant force pushes a block along a horizontal frictionless surface. The block moves from rest through a fixed distance.

Question 1

A constant force pushes a block along a horizontal frictionless surface. The block moves from rest through a fixed distance.

What is the relationship between the final speed v of the block and its mass m?

Reference: Past Exam Paper – June 2017 Paper 13 Q16

Solution:

Answer: C.

v² = u² + 2as

Since the block moves from rest, u = 0.

v² = 2as

So, v² ∝ a

Since a constant force is applied, F = ma, giving a = F/m

So, acceleration a ∝ 1/m

Replacing the relationship a ∝ 1/m into the equation involving velocity,

v² ∝ 1 / m

v ∝ 1 / √m

Variables x and y are related by the equation y = p – qx where p and q are constants.

Question 2

Variables x and y are related by the equation y = p – qx where p and q are constants.

Values of x and y are measured experimentally. The results contain a systematic error.

Which graph best represents these results?

Reference: Past Exam Paper – November 2012 Paper 12 Q6

Solution:

Answer: A.

Equation: y = p – qx  

Comparing with the equation of a line y = mx + c, the gradient is negative and constant (m = -q). So, it is a straight line. [C and D incorrect]

Systematic error is due to the instrument, so all points should lie on the line as the error always occurs in the same direction.

From the equation, when x = 0, y = p, but due to systematic error, this value is reduced.

Choice A is correct as it shows no random error but a systematic error with the line not going through p.

Choice B shows random error without systematic error.

An object hangs by means of two cords around two rods, as shown. The object is held in equilibrium by the forces F1 and F2.

Question 1

An object hangs by means of two cords around two rods, as shown.

The object is held in equilibrium by the forces F1 and F2. The object weighs 10 N. There is negligible friction between the rods and cords. Angle θ is 90°.

Which row of the table gives an angle θ of 90°?

F1 / N               F2 / N

A          4.0                   6.0

B          6.0                   4.0

C         6.0                   8.0

D         8.0                   6.0

Reference: Past Exam Paper – March 2017 Paper 12 Q14

Solution:

Answer: C.

The 2 forces F₁ and F₂ will also be along the cords.

Since the object is held in equilibrium, the resultant of the forces F₁ and F₂ should be equal to the weight of the object (= 10N).

The resultant of 3 vectors can be obtained by the triangle method.

From the question, we know that the angle between F₁ and F₂ is 90°. So, the triangle formed is actually a right-angled triangle. Also, the weight of the object is taken as the hypotenuse (since the other 2 forces formed the right angle).

From Pythagoras’ theorem,

F₁² + F₂² = 10²

It can be seen that both choices C and D apply to this equation.

But, now, let’s say that both forces were of the same magnitude, the object would have been midway between the 2 rods. But since the object is closer to the rod on the right, the force should be greater there.

So, F₂ > F₁.