A ball of mass 0.7Kg is dropped unto a floor from a height of 5m and rebounds to a height of 3m. Find the impulse on the ball when it hits the floor
impulse=changemomentum
= final momentum-initial
now, the ratio of velocities :
from height 5
mg5=1/2 m v^2
or v=sqrt (10g)
and v to get to height 3 m is
v=sqrt(6g)
impulse=changemomentum=m*(sqrt6g+sqrt3g)
1.5m
To find the impulse on the ball when it hits the floor, we need to use the concept of impulse-momentum theorem.
First, we need to understand the concept of impulse. Impulse is defined as the change in momentum of an object. It is calculated by multiplying the force applied to an object by the time interval over which the force is applied. Mathematically, it can be expressed as:
Impulse = Force x Time
The momentum of an object is given by the formula:
Momentum = Mass x Velocity
Now, let's break down the problem and calculate the impulse on the ball when it hits the floor.
Given:
Mass of the ball (m) = 0.7 kg
Initial height (h₁) = 5 m
Rebound height (h₂) = 3 m
To find the velocity just before the ball hits the floor, we can use the principle of conservation of mechanical energy. The initial potential energy of the ball (mgh₁) gets converted into kinetic energy just before hitting the floor.
Potential energy at height h₁ = Kinetic energy just before hitting the floor
mgh₁ = 0.5mv²
Simplifying this equation, we get:
v² = 2gh₁
Now, let's calculate the velocity (v) just before the ball hits the floor:
v² = 2 * 9.8 m/s² * 5 m
v² = 98 m²/s²
v ≈ 9.9 m/s (taking the positive value)
The velocity just before the ball hits the floor is approximately 9.9 m/s.
Next, we need to calculate the velocity of the ball just after bouncing back using the principle of conservation of mechanical energy again. The final potential energy of the ball (mgh₂) is converted into kinetic energy after rebounding.
Potential energy at height h₂ = Kinetic energy just after rebounding
mgh₂ = 0.5mvf²
Simplifying this equation, we get:
vf² = 2gh₂
Now, let's calculate the velocity (vf) just after the ball bounces:
vf² = 2 * 9.8 m/s² * 3 m
vf² = 58.8 m²/s²
vf ≈ 7.7 m/s (taking the positive value)
The velocity just after the ball rebounds is approximately 7.7 m/s.
Finally, we can calculate the impulse on the ball using the equation:
Impulse = Change in momentum = Final momentum - Initial momentum
Momentum is given by the formula:
Momentum = Mass x Velocity
Initial momentum (P₁) = m * v₁
Final momentum (P₂) = m * v₂
Impulse = P₂ - P₁
Impulse = m * v₂ - m * v₁
Impulse = m * (v₂ - v₁)
Plugging in the values, we have:
Impulse = 0.7 kg * (7.7 m/s - 9.9 m/s)
Impulse = 0.7 kg * (-2.2 m/s)
Impulse ≈ -1.54 Ns
The impulse on the ball when it hits the floor is approximately -1.54 Ns.