It is important to remember that a vector has both a magnitude and a direction, the first three options dont have a direction associated with them, it makes no sense to say 5 minutes East, whereas a displacement has a direction, it points from point A to point B.
CloseFirstly we should note that the \( x \) and \( y \) components of the velocity can be thought of independently, meaning they do not affect one another. The key thing to realize in this question is that air resistance can be ignored, meaning we will have no force/acceleration in the horizontal direction, and thus, no change in speed, hence answer \( B \).
CloseIf we look at the free body diagram of the falling ball, we see that the weight points down, whereas the drag force acts opposite to our velocity, so upwards. Since our ball is flying down, it is being slowed down by gravity, hence we know \( mg > cv \), so if we subtract \( cv \) from \( mg \), we will get our net force, which is \( F = ma \).
Close
Many IB questions require for us to notice key words in questions. This question is no different, to solve it, we must recognize that it is at constant speed, meaning the sum of the vertical forces on the cube must be zero, since there is no acceleration. For this to happen the resistive force must equal the weight, hence B.
CloseWhenever we need to examine the area/gradient of a graph, it is good practise to consider the units on the 2 axes. On the vertical axis we have \(\dfrac{m}{s}\) and the horizontal we have \(\text{s}\). When we are finding area, we multiply the lengths of graph on the x and y axes respectively, hence we also multiply the units present, so in this case if we do so we get meters \(\dfrac{m}{s} * {s} = {m}\), which leads us to think displacement. The same logic can be applied to the gradient, where we always divide the length of the y axis with m the x axis. So we have \(\dfrac{\dfrac{m}{s}}{s} = {\dfrac{m}{s^2}}\) , which is the acceleration.
CloseSince it is thrown up, it will have an initial y velocity, but gravity will be slowing it down, so \(\text{v}_{y}\) decreases. In this case air resistance is also present, and it acts opposite to our velocity, thus it is also decreasing our \(\text{v}_{x}\).
CloseWe can use conservation of energies, at the start we had potential and kinetic energy, at the end we only have the work done by the frictional force. So:
\[ mgh + \frac{1}{2}mv^2 = F_f \cdot d = N \cdot \mu \cdot d = mg \mu d \]
\[ \mu = \frac{gh + \frac{1}{2}mv^2}{gd} \]
CloseWe can use the formula \( v = u + at \), where \( 0 = 20 - 4t \), so \( t = 5 \).
CloseAt the start, resistive force is 0, so we have maximum acceleration, and the speed increases. However, as the speed increases, so does the resistive force, up until the resistive force equals the pulling force. Then, the acceleration will become 0, as the net force is also 0. Thus, the velocity will remain at a constant value.
CloseIn the \( x \) direction, we have air resistance acting, which is always trying to slow the ball down, hence reducing the \( x \) component. In the \( y \) direction, we have gravity increasing the ball's \( y \) component.
CloseMomentum is mass times velocity, where velocity is a vector, thus momentum is also a vector. Distance, speed, and electric potential are scalars.
CloseThere are 2 forces acting on the car, one being the weight force and the drag force.
So the net force is \( F = ma \), plugging in gives:
\[ 1500 \times 9.81 - 100 = 1500a \]
So \( a = \text{9.74 ms}^{-2} \).
CloseFirst we need to calculate the time needed for the ball to drop, which we can do with the equation \( s = ut + \frac{1}{2} a t^2 \), plugging in \( s = 100 \) m and \( u = 0 \), as the initial speed in the y direction was 0, we get \( t = 4.515 \) s.
Since we assume air resistance is negligible, the \( x \)-component must be constant. Thus, we can simply use \( d = v t = 10 \times 4.515 = 45.15 \) m.
CloseWe can see Joel's mass is \( \frac{500}{9.81} = 51.02 \) kg. We also know that the weight measured by the scale is the normal force exerted by Joel on the scale. Thus, the two forces acting on Joel are his weight and the normal force. So we can write the net force as \( F_{\text{net}} = ma = N - W \), so \( a = \frac{N - W}{m} = \text{-3.92 ms}^{-2} \) (downwards).
CloseWe can use the equation \( s = ut + \frac{1}{2}at^2 \), where we know \( u = 0 \), as the player did not give the ball any vertical velocity. So, by plugging into the formula, we get \( s = \frac{1}{2} \cdot 9.81 \cdot 0.714^2 = 2.5 \, \text{m} \).
Close