A pendulum with a bob of mass 30g is lifted to height of 5.00cm above its lowest point and allowed to fall so that collides elastically with a block of mass 20g that rests on a smooth horizontal floor. The block stops 5 seconds after the collision. Assuming no sound is generated during the collision, determine the:

a)speed of the bob just before it collides with the block
b)distance travelled by the block before it stops
c)deceleration of the block

PE=KE

m₁gh =m₁v₁²/2
v₁ =sqrt(2gh) = sqrt(2•0.03•0.05) =
=0.055 m/s.

m₁v₁ = m₁u₁ + m₂u₂
u₂=2m₁v₁/(m₁+m₂)=2•0.03•0.055/(0.03+0.02) = 0.066 m/s.

0= u₂ -at
a= u₂/t=0.066/5= 0.0132 m/s²,

s=at²/2=0.0132•5²/2=0.165 m

To solve this problem, we can apply the principles of conservation of mechanical energy and momentum.

a) To determine the speed of the bob just before it collides with the block, we will equate the potential energy of the bob at its highest point with its kinetic energy just before the collision.

The potential energy of the bob at its highest point can be calculated using the formula:

Potential energy (PE) = m * g * h

where m is the mass (30g = 0.03 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (5.00 cm = 0.05 m).

PE = 0.03 kg * 9.8 m/s² * 0.05 m
PE = 0.0147 J

At the highest point, all the potential energy is converted into kinetic energy, so:

Kinetic energy (KE) = 0.0147 J

Using the formula for kinetic energy:

KE = (1/2) * m * v^2

where m is the mass of the bob (0.03 kg) and v is the velocity just before the collision.

0.0147 J = (1/2) * 0.03 kg * v^2
0.0147 J = 0.015 kg * v^2
v^2 = 0.0147 J / 0.015 kg
v^2 = 0.98 m²/s²
v = √(0.98 m²/s²)
v ≈ 0.99 m/s

Therefore, the speed of the bob just before it collides with the block is approximately 0.99 m/s.

b) To determine the distance traveled by the block before it stops, we will use the equation:

Distance (d) = Initial velocity (u) * time (t) + (1/2) * acceleration (a) * t^2

The initial velocity of the block is the same as the velocity of the bob just before the collision, which is approximately 0.99 m/s.

Using the equation for distance:

Distance (d) = 0.99 m/s * 5 s + (1/2) * 0 * (5 s)^2
Distance (d) = 4.95 m

Therefore, the block travels approximately 4.95 meters before it stops.

c) The deceleration of the block can be determined using the equation:

Deceleration (a) = Change in velocity (Δv) / Time (t)

The change in velocity is the initial velocity (0.99 m/s) minus the final velocity (0 m/s) when the block stops.

Deceleration (a) = (0 - 0.99 m/s) / 5 s
Deceleration (a) = -0.99 m/s / 5 s
Deceleration (a) ≈ -0.198 m/s²

Therefore, the deceleration of the block is approximately -0.198 m/s². The negative sign indicates it is decelerating or slowing down.

To solve this problem, we can use the principle of conservation of mechanical energy and Newton's laws of motion. Let's break it down into three parts:

a) Speed of the bob just before it collides with the block:
1. First, we need to find the potential energy of the pendulum bob at its highest point.
Potential energy = mass * gravity * height
Potential energy = 0.03 kg * 9.8 m/s^2 * 0.05 m
Potential energy = 0.0147 J

2. Considering the collision is elastic, the total mechanical energy of the system (pendulum bob + block) is conserved.
Total mechanical energy before collision = Total mechanical energy after collision

3. At its highest point, the pendulum bob has only potential energy. Just before the collision, it has kinetic energy.
So, the kinetic energy just before the collision = Potential energy at the highest point.
Kinetic energy = 0.0147 J

4. Kinetic energy (KE) can be represented as 1/2 * mass * speed^2.
0.0147 J = 0.03 kg * (speed)^2
speed^2 = 0.0147 J / 0.03 kg
speed^2 = 0.49 m^2/s^2
speed ≈ 0.7 m/s

Therefore, the speed of the bob just before it collides with the block is approximately 0.7 m/s.

b) Distance traveled by the block before it stops:
1. Using Newton's second law of motion, we know that force (F) is equal to mass (m) multiplied by acceleration (a).
F = m * a

2. The only force acting on the block is its weight (mg). Since it comes to rest, the net force acting on it is zero.
F = 0

3. Therefore, acceleration (a) is also zero. This means the block is moving at a constant speed after the collision, covering no further distance.

So, the distance traveled by the block before it stops is zero.

c) Deceleration of the block:
Since the block comes to rest, its velocity changes from a non-zero value to zero over a time interval of 5 seconds.
Deceleration = (final velocity - initial velocity) / time
Deceleration = (0 - initial velocity) / 5

As we calculated earlier, the initial velocity (speed of the bob just before collision) is approximately 0.7 m/s.
Deceleration = (0 - 0.7 m/s) / 5
Deceleration ≈ -0.14 m/s^2

So, deceleration of the block is approximately -0.14 m/s^2. The negative sign indicates deceleration or slowing down.