Particle moving through liquid or resistance

A question is this type if and only if a particle falls through air and then enters a liquid (or experiences a resistance force), changing acceleration, and the task involves finding the new acceleration, resistance force, or time/distance in the liquid using SUVAT in each phase.

3 questions · Moderate -0.1

3.02d Constant acceleration: SUVAT formulae
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CAIE M1 2006 June Q4
8 marks Standard +0.3
4 \includegraphics[max width=\textwidth, alt={}, center]{b5873699-d207-4cad-9518-1321dc429c15-3_568_1084_269_532} The diagram shows the velocity-time graph for the motion of a small stone which falls vertically from rest at a point \(A\) above the surface of liquid in a container. The downward velocity of the stone \(t \mathrm {~s}\) after leaving \(A\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The stone hits the surface of the liquid with velocity \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when \(t = 0.7\). It reaches the bottom of the container with velocity \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when \(t = 1.2\).
  1. Find
    1. the height of \(A\) above the surface of the liquid,
    2. the depth of liquid in the container.
    3. Find the deceleration of the stone while it is moving in the liquid.
    4. Given that the resistance to motion of the stone while it is moving in the liquid has magnitude 0.7 N , find the mass of the stone.
CAIE M1 2014 November Q6
9 marks Standard +0.3
6 A particle of mass 3 kg falls from rest at a point 5 m above the surface of a liquid which is in a container. There is no instantaneous change in speed of the particle as it enters the liquid. The depth of the liquid in the container is 4 m . The downward acceleration of the particle while it is moving in the liquid is \(5.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
  1. Find the resistance to motion of the particle while it is moving in the liquid.
  2. Sketch the velocity-time graph for the motion of the particle, from the time it starts to move until the time it reaches the bottom of the container. Show on your sketch the velocity and the time when the particle enters the liquid, and when the particle reaches the bottom of the container.
OCR MEI M1 2005 January Q6
19 marks Moderate -0.8
6 In this question take \(g\) as \(10 \mathrm {~m \mathrm {~s} ^ { - 2 }\).} A small ball is released from rest. It falls for 2 seconds and is then brought to rest over the next 5 seconds. This motion is modelled in the speed-time graph Fig. 6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c84a748a-a6f4-48c5-b864-fe543569bdf5-5_659_1105_578_493} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} For this model,
  1. calculate the distance fallen from \(t = 0\) to \(t = 7\),
  2. find the acceleration of the ball from \(t = 2\) to \(t = 6\), specifying the direction,
  3. obtain an expression in terms of \(t\) for the downward speed of the ball from \(t = 2\) to \(t = 6\),
  4. state the assumption that has been made about the resistance to motion from \(t = 0\) to \(t = 2\). The part of the motion from \(t = 2\) to \(t = 7\) is now modelled by \(v = - \frac { 3 } { 2 } t ^ { 2 } + \frac { 19 } { 2 } t + 7\).
  5. Verify that \(v\) agrees with the values given in Fig. 6 at \(t = 2 , t = 6\) and \(t = 7\).
  6. Calculate the distance fallen from \(t = 2\) to \(t = 7\) according to this model.