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Thankyou so much man! Thanks alot!!For Question 13, there are a few important points to be noticed in the question:
i) the string is light, so we can assume that the magnitude of the tension for in the string is the same at all points along the string;
Suppose the string were indeed heavy, each part of the string would have a particular mass, and since the string itself is accelerating, there would have to be a net force on each one of those parts. However, the only forces on the string would be gravity and the 2 tension forces, one from each side. So, for a horizontal part of the string, the tension would have to be different at each point (otherwise no individual part could accelerate, and so the rope on a whole won't accelerate either.
ii)all surfaces is frictionless, so we need not make any assumptions concerning the magnitude of any friction forces and so this simplifies the question a lot.
Now onto Newton's Laws:
Taking the 2.0 kg mass into account, we need only look at the horizontal forces , since there is no vertical acceleration of that mass.
The only horizontal force acting on the mass is the Tension in the string (which, as noted before, is constant throughout the length of the string).
Therefore, T = (2.0) * a
In other words, our positive direction for the 2.0 kg mass is towards the right, and the tension force is positive since it acts towards the right.
Taking the 1.0 kg mass into account, we need only look at the vertical forces, since there is no horizontal/ sideways acceleration of the 1.0 kg mass.
The only 2 horizontal forces acting on the mass are the Tension in the string (same value as in the previous equation) and it's weight (= 1.0 * 9.81 = 9.81 Newtons).
Therefore, since Tension acts in the direction opposite to the acceleration and the weight acts in the direction of the acceleration,
-T + 9.81 = (1.0) * a
In other words, for the 1.0 kg mass, our positive direction is downwards. Therefore, the tension is negative and the weight in positive (in terms of sign value).
The acceleration of each object is the same, since the string can be assumed to be rigid and in-extensible (cannot be stretched), so any movement of one mass will the mirrored in the movement of the other mass, the only difference being the direction.
Solving these equations, we get a = 9.81 / 3 = 3.27 ms^-2
Since this acceleration is constant, we can apply the constant acceleration equations for either object. Suppose we take the 2.0 kg mass, it will have traveled 0.5 meters when the other object hits the ground, it started from rest, and it moves with the acceleration calculated above, so
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.5 m
v = ?
v^2 = u^2 + 2 * a * s
v^2 = 0 +3.27
So, the velocity v = 1.81 ms^-1 = A
For the next question, we can use several theories and laws, but lack of available information constricts our choices like so:
i) We can use Newton's Laws, but we do not know what the magnitudes of the forces that act between the two objects during the collision are; we also do not know how long the collision lasts, so we cannot answer this question using Newton's Laws.
ii) We can use the Work-Energy theorem, but there too we don't know the magnitudes of the forces acting during the collision and we do not know the distance the object move during the collision. So, we cannot use the Work-Energy theorem either.
iii) We can use the Principle of Conservation of Momentum, which does allow us to answer this question; the reason for this is that we do not need to know the forces acting between the objects involved if we take them both as our system.
So, taking both the pellet and the block as the system, we have:
Initial Momentum = (200 ms^-1) * (0.005 kg) = 1 kg ms^-1
Since no external forces act on the system during the collision (gravity is a rather negligible force compared to the forces acting between the two objects; it does cause some change, but gravity has rather little significance during the collision), momentum is assumed to be conserved. Not only that, since the pellet is embedded in the block, the two objects move as one mass, with a common velocity.
Therefore,
Final Momentum = Initial Momentum = [(0.095 + 0.005) kg] * [(final velocity) ms^-1)
0.1 * final velocity = 1
Therefore, final velocity = 10 ms^-1
During the collision we can assume that the objects don't move much upwards, and only start moving at 10 ms^-1 after the collision is complete. Therefore, since the only force acting on the system of the two objects is gravity = -0.981 Newtons, we can use the equations of constant acceleration motion:
u = 10 ms^-1
v = 0 ms^-1 (at the top of it's ascent)
a = -9.81 ms^-2
so v^2 = u^2 + 2as
0 = 100 - 2 * 9.81 * s
Therefore, s = 5.1 meters = A
Hope this helps!
Best of luck for your exams!
