Impulsive Force— a large force acting for a short period of time
The impulse of a force (I) is defined as the product of the force and the time during
which it acts.
$$ I = F \times t $$
Where:
I → impulse (N·s)
F → force (N)
t → time in seconds
Only external forces give impulses to a body. The direction of the impulse is the same as the direction of the force involved.
Momentum is a quantity of motion. It is the property of a body that determines the time required to bring it to rest.
It is the product of its mass m and instantaneous velocity v.
$$ \text{Momentum} = m \times v $$
5. The unit of momentum derived from mass and velocity is kg·m/s.
Conservation of Momentum
If two bodies collide, then the sum of the momenta before collision is equal to the sum of the momenta after collision.
$$ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' $$
Where:
m1 → mass of the first body
m2 → mass of the second body
v1 → velocity of the first body before collision
v2 → velocity of the second body before collision
v1' → velocity of the first body after collision
v2' → velocity of the second body after collision
Sign Convention for Velocities:
v is positive (+) if its direction is to the right (before or after collision).
v is negative (−) if its direction is to the left (before or after collision).
$$ m v_1 + F_{\text{pos}} \, t \; - \; F_{\text{neg}} \, t = m v_2 $$
Coefficient of Restitution
The relative velocity of two bodies before collision is v1 − v2
and the relative velocity after collision is v1′ − v2′.
The coefficient of restitution (e) is the ratio of the relative velocity after
impact to that before impact.
$$ e = \frac{(v_2' - v_1')}{(v_1 - v_2)} $$
For perfectly inelastic bodies, e = 0
For perfectly elastic bodies, e = 1
Problem: Work and Energy Method & Impulse and Momentum
A tensile force of 50N inclined 30° from the horizontal is applied on a 200N block which is at rest.
a. Determine the coefficient of friction between the block and the horizontal plane if after it moves a distance of 40 meters, its velocity is 10m/s.
b. Compute the distance (meters) traveled by the block at the end of 20 seconds.
Problem: Sphere Swinging down to Hit a Box
The 2-kg sphere is released from rest and swings down, striking the 5-kg box B. The distance from the ceiling to the center of the sphere is 1 meter. The coefficient of restitution for the collision is 0.70 and the coefficient of friction between the box and the floor is 0.10. If the box slides 750mm after the impact before it comes to rest again,
a. Determine the velocity (in m/s) of the box immediately after the impact.
b. Determine the angle 𝜙 at which the sphere A was released.
Problem: Sphere Swinging down to Hit a Box
Object A approaches object B which is stationary. The velocity of object A is V=28i+5j-3k in m/s. Mass of A =10kg and that of B =4kg. If the impact is perfectly inelastic,
a. Compute the x-component of the velocity after impact.
b. Compute the y-component of the velocity after impact.
c. Compute the z-component of the velocity after impact.
