Question 1

An object moves from rest in a straight line with a uniform acceleration and takes 5 s to cover the first 100 m. It then continues with the same acceleration for another 100 m. The velocity after the total distance of 200 m will be:

A) 40.0 m/s
B) 56.6 m/s
C) 80.0 m/s
D) 160 m/s

Accepted Answer

Using the formula for uniform acceleration:

distance = 1/2 × acceleration × time²

We can find the acceleration:

100 = 1/2 × acceleration × 5²

Acceleration = 8 m/s²

Then we can use the formula for velocity with uniform acceleration:

final velocity = initial velocity + acceleration × time

For the first 100 m:

final velocity = 0 + 8 × 5 = 40 m/s

For the second 100 m:

final velocity = 40 + 8 × 5 = 80 m/s

Therefore, the best choice is B, 40 m/s.

Question 2

An object moves from rest in a straight line with a uniform acceleration and takes 5 s to cover the first 100 m. It then continues with the same acceleration for another 100 m. If the object continues with the same acceleration, the time taken to cover the next 100 m will be:

A) 2.1 s
B) 5.0 s
C) 7.1 s
D) 10.0 s

A

Accepted Answer

Explanation:The object covers the first 100 m in 5 s, so its acceleration is a = 20 m/s^2. The time taken to cover the next 100 m will be t = sqrt(2s/a) = sqrt(2 * 100 m / 20 m/s^2) = 2.1 s.

Question 3

An object moves from rest in a straight line with a uniform acceleration and takes 5 s to cover the first 100 m. It then continues with the same acceleration for another 100 m. The velocity after the total distance of 200 m will be:

A) 40.0 m/s
B) 56.6 m/s
C) 80.0 m/s
D) 160 m/s

B

Accepted Answer

The object moves with uniform acceleration, so we can use the equations of motion. For the first 100 m, we have $s = ut + \frac{1}{2}at^2$, where $s$ is the distance, $u$ is the initial velocity (0 m/s since it starts from rest), $a$ is the acceleration, and $t$ is the time. Plugging in the values, we get $100 = 0 + \frac{1}{2}a(5^2)$, which simplifies to $a = 8 \, \text{m/s}^2$.To find the velocity after 200 m, we use the equation $v^2 = u^2 + 2as$, where $v$ is the final velocity, $u$ is the initial velocity (0 m/s), $a$ is the acceleration (8 m/s$^2$), and $s$ is the total distance (200 m). Plugging in the values, we get $v^2 = 0 + 2(8)(200)$, which simplifies to $v^2 = 3200$, and taking the square root gives $v = 56.6 \, \text{m/s}$.

Question 4

An object falls freely from rest. If it falls through a distance s in the first second, the distance it will fall through in the second second is:&#10;A) 1s&#10;B) 2s&#10;C) 3s&#10;D) 4s

Accepted Answer

The distance an object falls due to gravity increases proportionally with the square of the time elapsed. If it falls a distance s in the first second, in the next second (the second second), it will fall through an additional 3s, making the total distance for the second second alone 3s. This is because the distance fallen is proportional to the square of the time, so in 2 seconds, it falls 2^2 = 4 times the distance it falls in 1 second, meaning the additional distance fallen in the second second is 3 times the distance fallen in the first second.

Question 5

Decide whether each of these statements is True (T) or False
For a velocity−time graph:
(i) The distance travelled between two times is the area under the graph between the ordinates for those times.
(ii) The acceleration at any instant is the gradient of the graph at that time.

A) (i) T (ii) T
B) (i) T (ii) F
C) (i) F (ii) T
D) (i) F (ii) F

Accepted Answer

(i) True, because the area under a velocity-time graph represents the displacement (or distance if the velocity does not change direction) between two points in time. (ii) True, because the gradient (slope) of a velocity-time graph at any point gives the acceleration at that instant.

Question 6

The following is a force&#8722;extension graph for strips of materials A, B, C and D, each&#10;Having initially the same cross-sectional area and length. When the strip is used as a cantilever, which one will sag the most when force F is applied to the free end?&#10;A) A&#10;B) B&#10;C) C&#10;D) D

Accepted Answer

Material D will sag the most because it has the lowest Young's modulus, which means it is the least stiff and most flexible. The graph shows that at the same applied force, material D is able to extend the most, indicating that it will sag the most when used as a cantilever.

Question 7

For a material with a tensile modulus of elasticity of 200 GPa, the strain produced by a stress of 50 MPa, assuming the limit of proportionality is not exceeded, will be:&#10;A) 0.000 25&#10;B) 0.25&#10;C) 4&#10;D) 4000

Accepted Answer

The relationship between stress (σ), strain (ε), and modulus of elasticity (E) is given by the equation:

σ = Eε

We are given E = 200 GPa (or 200 x 10^9 Pa) and σ = 50 MPa (or 50 x 10^6 Pa).

Plugging these values into the equation, we get:

50 x 10^6 = 200 x 10^9 ε

Solving for ε, we get:

ε = (50/200) x 10^-3 = 0.00025

Therefore, the strain produced by a stress of 50 MPa is 0.00025, which corresponds to answer choice A.

Question 8

The maximum tensile force a rectangular steel bar of cross-section 20 mm % 10 mm can withstand, if it has a tensile strength of 800 MPa, is:&#10;A) 16 kN&#10;B) 25 kN&#10;C) 160 kN&#10;D) 2500 GN

Accepted Answer

The maximum tensile force that a bar can withstand can be calculated using the formula:
Maximum Tensile Force = Tensile Strength x Cross-sectional Area
Cross-sectional Area = 20 mm x 10 mm = 200 mm² = 0.0002 m²
Tensile Strength = 800 MPa = 800 N/mm² = 800,000,000 N/m²
Maximum Tensile Force = 800,000,000 x 0.0002 = 160,000 N = 160 kN
Therefore, the correct choice is C.

Question 9

A rectangular bar has a cross-sectional area of 40 mm2. The tensile stress produced when it is stretched by axial forces of 100 N is:&#10;A) 0.4 MPa&#10;B) 2.5 MPa&#10;C) 25 MPa&#10;D) 40 MPa

Accepted Answer

Tensile stress is calculated as the force applied divided by the cross-sectional area. Here, it is 100 N / 40 mm^2 = 2.5 N/mm^2, which is equivalent to 2.5 MPa.

Deck 2: Linear Motion, and Stress and Strain