Two particles A and B of mass 5kg and 9kg respectively rests on the smooth faces of a fixed wedge .They are connected by a light inextensible string passing over a smooth pulley at C and are released from rest.In the subsequent motion B hits the ground and does not rebound. a.) the speed of the particles when B hits the ground. b.)the acceleration of A after B hits the ground c.) the distance of A from C when A first comes to rest.
To answer these questions, we need to analyze the physical principles involved in this scenario. Let's break it down step by step:
Step 1: Analyzing forces and motion before B hits the ground
Before B hits the ground, both particles A and B are connected by a light inextensible string, which passes over a smooth pulley at C. Since A and B are connected by the string, they will always have the same acceleration.
Both particles are initially at rest, and there are no external forces acting on the system. However, there are internal forces acting within the system due to the tension in the string.
Let's consider positive acceleration as downward for both particles.
The force acting downwards on A is its weight (mg), where m is the mass of A (5 kg) and g is acceleration due to gravity (9.8 m/s^2).
The force acting upwards on B is the tension in the string (T).
Since the system is in equilibrium before any motion starts, these forces are equal in magnitude:
mg = T
Step 2: Analyzing motion after B hits the ground
Once B hits the ground, it will not rebound, which means the string becomes slack. This implies that the tension (T) in the string becomes zero.
Since the tension becomes zero, the only force acting on A is its weight (mg). Therefore, A will continue to accelerate vertically downwards under the influence of gravity after B hits the ground.
Now let's answer each of the given questions using the principles we have discussed:
a) The speed of the particles when B hits the ground:
To find the speed of the particles when B hits the ground, we need to determine the acceleration of the system using Newton's second law (F = ma) and then apply the equations of motion.
Since B hits the ground, the distance traveled by B is the same as the height of the wedge. Let's call this distance h.
Using the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (in this case, u = 0 as both particles are initially at rest)
a = acceleration (unknown)
s = distance traveled (h)
Substituting the values, we get:
v^2 = 0^2 + 2ah
v^2 = 2ah
To find the value of acceleration (a), we can use Newton's second law by considering the forces acting on the system before B hits the ground. The net force in the vertical direction is the weight of A (mg).
Using F = ma:
mg = ma
a = g
Now substituting this value of acceleration in the equation for v^2, we get:
v^2 = 2gh
v = sqrt(2gh)
Therefore, the speed of the particles when B hits the ground is v = sqrt(2gh).
b) The acceleration of A after B hits the ground:
As discussed earlier, A will continue to accelerate vertically downwards under the influence of gravity after B hits the ground. Therefore, the acceleration of A is equal to the acceleration due to gravity, which is g = 9.8 m/s^2.
c) The distance of A from C when A first comes to rest:
To find the distance of A from C when A first comes to rest, we need to consider the motion of A after B hits the ground.
Using the equations of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as A comes to rest)
u = initial velocity (unknown)
a = acceleration (g = 9.8 m/s^2)
s = distance traveled (d)
Since A comes to rest, the final velocity (v) is 0 m/s.
0^2 = u^2 + 2gd
u^2 = -2gd
u = sqrt(-2gd) (taking the positive value)
Therefore, the distance of A from C when A first comes to rest is given by the equation:
d = u^2/2g
d = (-2gd)/2g
d = -d
This means that the distance of A from C when A first comes to rest is equal to the distance traveled by B before hitting the ground (h).
To solve this problem, we'll use the principle of conservation of energy. Let's break it down into three different parts.
Part a) Finding the speed of the particles when B hits the ground:
1. Let h be the height at which B starts from.
2. The potential energy of B at height h is given by mgh, where mg is the gravitational force acting on B.
The potential energy of A at this height is 0 since it is on the smooth face of the wedge.
3. When B hits the ground, all of its potential energy is converted into kinetic energy.
The kinetic energy of B is given by (1/2)mv^2, where v is the speed of B.
The kinetic energy of A at this point is also (1/2)mv^2 since the string is inextensible and the particles are connected.
4. Equating the potential and kinetic energy:
mgh = (1/2)mv^2 + (1/2)mv^2
mgh = mv^2
v^2 = gh
v = √(gh)
Substituting the values of mass and acceleration due to gravity:
v = √(9.8 * h)
Part b) Finding the acceleration of A after B hits the ground:
1. After B hits the ground, the only force acting on the system is the tension in the string.
2. The net force acting on A is equal to the tension, T, and is given by ma.
Therefore, T = ma.
3. The force acting on B is the tension, T, and the weight, mg, which is equal to 9g.
Therefore, T + 9g = 9a, since the mass of B is 9 kg.
4. We can now solve the simultaneous equations T = ma and T + 9g = 9a to find the value of a.
Part c) Finding the distance of A from C when A first comes to rest:
1. When A comes to rest, the tension in the string is zero.
2. The net force acting on A is equal to the weight, mg, and is given by ma.
Therefore, mg = ma.
3. Rearranging the equation, we get a = g.
4. Using the equation of motion s = ut + (1/2)at^2, where s is the distance, we have:
s = 0*t + (1/2)g*t^2
s = (1/2)g*t^2
t = √(2s/g)
These are the step-by-step solutions to the given problem.