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PHYSICS, M1 EQ-Bank 15

A table is slanted so a book resting on that table begins to accelerate at a constant value, \(a\).

If the book travels 45 cm in 1.0 seconds, determine its acceleration, \(a\)?   (2 marks)

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\(0.9\ \text{ms}^{-2}\ \ \text{down slope}\)

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\(u= 0\ \text{ms}^{-1}, \ t=1\ \text{s}, \ s = \dfrac{45}{100} = 0.45\ \text{m}\)

\(s\) \(=ut +\dfrac{1}{2}at^2\)  
\(0.45\) \(=0 + 0.5 \times a \times 1^2\)  
\(a\) \(=\dfrac{0.45}{0.5 \times 1^2}\)  
  \(=0.9\ \text{ms}^{-2}\ \ \text{down slope}\)  

PHYSICS, M1 EQ-Bank 13

Outline an experimental procedure to determine the acceleration of a falling steel ball. Your explanation should include all the measurements that must be recorded, the calculations needed to compute the acceleration, and an identification of any potential sources of error in the experiment.   (6 marks)

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Procedure and Measurements:

  • Set up a vertical drop area with a height of two metres which can be measured using a measuring tape or ruler.
  • Position the steel ball at the measured height using a release mechanism or having someone drop it from the height of 2 metres from rest.
  • Start the stopwatch (or begin video recording) at the moment of release and stop the timer as soon as the ball hits the ground.
  • Repeat the drop several times (e.g., 3–5 trials) to obtain an average time of fall (\(t\)).
  • The acceleration of the ball can be calculated using the formula, \(s= ut + \dfrac{1}{2}at^2\), where \(u=0\ \text{ms}^{-1}\), \(s=2\ \text{m}\) and \(t\) is the time measured for the ball to drop. Rearranging the formula,  \(a= \dfrac{2s}{t^2}\).

Sources of Error:

  • Timing: Not starting the stopwatch at the exact times when the ball is released or stopping the stopwatch at the exact time when the ball hits the ground. 
  • Initial Velocity: if the person holding the steel ball does not drop it from rest.
  • Height: Inaccuracies in the measurement of the two metre drop height.
Show Worked Solution

Procedure and Measurements:

  • Set up a vertical drop area with a height of two metres which can be measured using a measuring tape or ruler.
  • Position the steel ball at the measured height using a release mechanism or having someone drop it from the height of 2 metres from rest.
  • Start the stopwatch (or begin video recording) at the moment of release and stop the timer as soon as the ball hits the ground.
  • Repeat the drop several times (e.g., 3–5 trials) to obtain an average time of fall (\(t\)).
  • The acceleration of the ball can be calculated using the formula, \(s= ut + \dfrac{1}{2}at^2\), where \(u=0\ \text{ms}^{-1}\), \(s=2\ \text{m}\) and \(t\) is the time measured for the ball to drop. Rearranging the formula,  \(a= \dfrac{2s}{t^2}\).

Sources of Error:

  • Timing: Not starting the stopwatch at the exact times when the ball is released or stopping the stopwatch at the exact time when the ball hits the ground. 
  • Initial Velocity: if the person holding the steel ball does not drop it from rest.
  • Height: Inaccuracies in the measurement of the two metre drop height.

PHYSICS, M1 EQ-Bank 12

A skydiver jumps from a stationary aircraft and yells as soon as she starts falling. Four seconds later, while still falling, she hears the echo of her shout from the ground below.

  1. Calculate how far the skydiver has fallen after 4 seconds.   (1 mark)
  2. What is the total distance the sound of their shout travelled in 4 seconds? (Hint: the speed of sound is 340 m/s)   (1 mark)
  3. Determine the altitude from which the skydiver jumped if this scenario occurred.   (2 marks)
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a.    \(78.4\ \text{m}\)

b.    \(1360\ \text{m}\)

c.    \(719.2\ \text{m}\)

Show Worked Solution

a.    \(u=0,\ \ t=4,\ \ a=9.8\)

Using the formula  \(s=ut + \dfrac{1}{2}at^2\)

\(s = (0 \times 4) + (\dfrac{1}{2} \times 9.8 \times 4^2) = 78.4\ \text{m}\)
 

b.    \(d=vt = 340 \times 4 = 1360\ \text{m}\)
 

c.    Let initial height of the skydiver be \(h\ \text{m}\).

  • When the skydiver hears the echo of their scream they would be at a height of \((h-78.4)\ \text{m}\).
  • As the sound of the scream travelled both of those distances:
\(h+h-78.4\) \(=1360\)  
\(2h\) \(=1438.4\)  
\(h\) \(=719.2\ \text{m}\)  

PHYSICS, M1 EQ-Bank 3

A hot-air balloon is travelling at a constant upwards velocity of 15 ms\(^{-1}\).

A passenger on the hot-air balloon decides to time how long it takes a pen to hit the ground when dropped from a height of 50 m.
 

Ignoring air resistance, determine how long it will take the pen to hit the ground.   (4 marks)

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\(5.07\ \text{s}\)

Show Worked Solution

\(s= 50, \ a=9.8, \ u=-15\)

Downwards  \(\Rightarrow\)  positive direction

\(s\) \(=ut+\dfrac{1}{2}at^2\)  
\(50\) \(=-15t + \dfrac{1}{2} \times 9.8 \times t^2\)  
\(0\) \(=4.9t^2-15t-50\)  
\(t\) \(=\dfrac{15\pm \sqrt{225-4 \times 9.8 \times -50}}{2 \times 4.9}\)  
  \(=5.07, -2.01\)  

 

Time for the pen to fall = 5.07 seconds \( (t \gt 0) \).

PHYSICS, M1 EQ-Bank 2

A cricket ball is hit vertically upwards from ground level, it gains 70 metres vertically and then falls back to the ground.

At what time(s) will the ball be 40 metres above the ground? Ignore any air resistance, giving your answer in seconds, correct to two decimal places.  (4 marks)

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\(6.25\ \text{s}\ \ \text{and}\ \ 1.31\ \text{s}\)

Show Worked Solution

\(\text{At 70 m}\ \Rightarrow \ v=0\ \text{ms}^{-1}:\)

\(v^2\) \(=u^2+2as\)  
\(0^2\) \(=u^2+2 \times -9.8 \times 70\)  
\(u^2\) \(=1372\)  
\(u\) \(=37.04\ \text{ms}^{-1}\)  

 
\(\text{Find}\ t\ \text{when}\ \ s=40:\)

\(s\) \(=ut+\dfrac{1}{2}at^2\)
\(40\) \(=37.04t+\dfrac{1}{2} \times -9.8 \times t^2\)
\(4.9t^2\) \(-37.04t+40=0\)
\(t\) \(=\dfrac{37.04 \pm \sqrt{(-37.04)^2-4 \times 4.9 \times 40}}{2 \times 4.9}\)
  \(=6.25\ \text{s}\ \ \text{and}\ \ 1.31\ \text{s}\)