The momentum principle states that net force changes the the momentum of an object where momentum is the product of the MASS of the body and it's VELOCITY.

Thus, momentum = mass(kg) ×velocity( m/s)= kgm/s ( derived unit)

Also to determine the rate of change of momentum;

Mass=m, initial velocity =u, final velocity = V and time =t

Initial momentum = mu

Final momentum = mv

Change in momentum = mv - mu

= m( v- u)

Rate of change of momentum=

m( v -u)/t --> equation 1

But V = u + at

Hence, a = v -u/t

Substitute for v -u/ t in equation 1

F = Kma, where k is constant and has the value of 1.

Therefore F = ma

Newton is the SI unit if force. It is defined as the force which gives a mass of 1kg an acceleration of 1 m/s².

IMPULSE is defined as the change of momentum of a body (mv-mu) or the product of force and time.

Thus impulse = force × time = m (v-u)

The unit is Ns ( Newton seconds) which is the same as change in momentum.

3 0

2 years ago

A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ = (v² - u²)/2a

h₂ = h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ = 1.3 m + 0.54 m

h₂ = 1.84 m

7 0

2 years ago

In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to

dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, $a=\dfrac{v-u}{t}$ ....(1)

The third equation of kinematics is as follows :

$v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s$

Hence, it will take 25 seconds to increase the speed.

6 0

2 years ago