**A.1 Kinematics SL Paper 1 : Exam Style Questions**

## IBDP Physics SL Paper 1 – All Chapters

Topic: A.1 Kinematics SL Paper 1

Syllabus: Displacement, Velocity, Acceleration, Uniform Acceleration Equations, Projectile Motion

#### Questions-**A.1 Kinematics SL Paper 1 **

A block of mass \(2.0 \mathrm{~kg}\) accelerates uniformly at a rate of \(1.0 \mathrm{~m} \mathrm{~s}^{-2}\) when a force of \(4.0 \mathrm{~N}\) acts on it. The force is doubled while resistive forces stay the same. What is the block’s acceleration?

A. \(4.0 \mathrm{~ms}^{-2}\)

B. \(3.0 \mathrm{~m} \mathrm{~s}^{-2}\)

C. \(2.0 \mathrm{~m} \mathrm{~s}^{-2}\)

D. \( 1.0 \mathrm{~ms}^{-2}\)

**▶️Answer/Explanation**

Ans:B

We can use Newton’s second law to find the initial resistive force (\(R_1\)) and the final resistive force (\(R_2\)):

1. For the initial case (\(F_1\)), the net force (\(F_{\text{net}_1}\)) is responsible for the initial acceleration (\(a_1\)):

\[F_{\text{net}_1} = F_1 – R_1 = m \cdot a_1\]

Solving for \(R_1\):

\[R_1 = F_1 – m \cdot a_1\]

\[R_1 = 4.0 \, \mathrm{N} – (2.0 \, \mathrm{kg} \cdot 1.0 \, \mathrm{m/s^2}) = 4.0 \, \mathrm{N} – 2.0 \, \mathrm{N} = 2.0 \, \mathrm{N}\]

So, in the initial case, the resistive force (\(R_1\)) is \(2.0 \, \mathrm{N}\).

2. For the final case (\(F_2\)), the net force (\(F_{\text{net}_2}\)) is responsible for the final acceleration (\(a_2\)):

\[F_{\text{net}_2} \Rightarrow F_2 – R_1 = m \cdot a_2\]

Solving for \(a_2\):

\[2.0 \, \mathrm{kg}\cdot a_2 = 8.0 \, \mathrm{N} – 2.0 \, \mathrm{kg} \]

\[6.0 \, \mathrm{N} = 2.0 \, \mathrm{kg} \cdot a_2\]

\[a_2 = 3.0 \, \mathrm{m/s^2}\]

#### Question

A car travels clockwise around a circular track of radius R. What is the magnitude of displacement from X to Y?

A. \( R \frac{3 \pi}{2}\)

B. \(R \frac{\pi}{2}\)

C. \(R \sqrt{2}\)

D. \(R\)

**▶️Answer/Explanation**

Ans:C

The displacement from point X to point Y is a chord of the circle and can be represented by the line connecting points X and Y.

#### Question

A car accelerates uniformly. The car passes point \(X\) at time \(t_1\) with velocity \(v_1\) and point \(Y\) at time \(t_2\) with velocity \(v_2\). The distance \(X Y\) is \(s\).

The following expressions are proposed for the magnitude of its acceleration a:

I. \(a=\frac{2 s}{\left(t_2-t_1\right)^2}\)

II. \(a=\frac{v_2^2-v_1^2}{2 s}\)

III. \(a=\frac{v_2-v_1}{t_2-t_1}\)

Which is correct?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

**▶️Answer/Explanation**

Ans:C

To determine which expressions for the magnitude of acceleration (\(a\)) are correct, we can analyze each of the proposed expressions one by one:

I. \(a = \frac{2s}{(t_2 – t_1)^2}\)

Expression I is incorrect because it assumes that the initial velocity \(\left(v_1\right)\) is zero, which may not be the case for the uniformly accelerating car.

II. \(a = \frac{v_2^2 – v_1^2}{2s}\)

This expression is also correct. It’s derived from the kinematic equation for uniformly accelerated motion: \(v^2 = u^2 + 2as\), where \(v_2\) is the final velocity, \(v_1\) is the initial velocity, \(s\) is the displacement, and \(a\) is the acceleration. Solving for \(a\) results in this expression.

III. \(a = \frac{v_2 – v_1}{t_2 – t_1}\)

This expression is also correct and is derived from the definition of acceleration as the rate of change of velocity over time.

So, expressions (II, and III) are correct.

*Question*

P and Q leave the same point, travelling in the same direction. The graphs show the variation with time *t *of velocity *v *for both P and Q.

What is the distance between P and Q when *t *= 8.0 s?

A 20 m

B 40 m

C 60 m

D 120 m

**Answer/Explanation**

Ans: B

Distance travelled by P\(=\frac{1}{2}(2)\times 10+(5\times 20)\)

\(=20+100\rightarrow 120\)

Distance travelled by \(Q=\frac{1}{2}(8)\times 40\)

\(=160\)

so , the difference \(= 160- 120 \rightarrow 40\)

#### Question

The velocity–time graph for an accelerating object that is traveling in a straight line is shown below.

Which of the following is the change in displacement of the object in the first 5.0 seconds?

A. 25.0 m

B. 12.5 m

C. 5.0 m

D. 1.0 m

**Answer/Explanation**

## Markscheme

B

The area under the v- t graph will give the displacement .

\(Area = \frac{1}{2}\times 5\times 5\)

\(= 12.5m\)