A particle of mass 2kg moves along the curve defined by r=(4t^2-t^3)i-5tj+(t^4-2)k. At t=1 find (a) The momentum (b) The force acting on it
5
To find the momentum of the particle at t=1, we can start by finding its velocity vector at that moment. The velocity vector is calculated by taking the derivative of the position vector with respect to time.
Given the position vector r = (4t^2 - t^3)i - 5tj + (t^4 - 2)k, we can find the derivatives of each component:
r' = (d/dt)(4t^2 - t^3)i - (d/dt)(5t)j + (d/dt)(t^4 - 2)k
Taking the derivatives:
r' = (8t - 3t^2)i - 5j + (4t^3)k
Now, substitute t=1 into the expression for r':
r'(t=1) = (8(1) - 3(1)^2)i - 5j + (4(1)^3)k
= 8i - 5j + 4k
The velocity vector at t=1 is 8i - 5j + 4k.
To find the momentum, we multiply the mass of the particle by its velocity vector:
Momentum = mass * velocity
Given that the mass of the particle is 2kg, the momentum at t=1 is:
Momentum = 2 * (8i - 5j + 4k)
= 16i - 10j + 8k
Therefore, the momentum of the particle at t=1 is 16i - 10j + 8k.
Moving on to part (b), to calculate the force acting on the particle at t=1, we apply Newton's second law which states that the force acting on an object is equal to the rate of change of its momentum.
Force = d/dt (Momentum)
Since the mass of the particle is constant, the derivative of its momentum with respect to time is simply the derivative of its velocity vector:
Force = (d/dt) (8i - 5j + 4k)
= 0i - 0j + 0k
The force acting on the particle at t=1 is therefore zero.
Note: It's important to note that this solution assumes that the particle is in free motion and is not subjected to any external forces.
To find the momentum and force acting on the particle at t = 1, we need to differentiate its position vector, r(t), with respect to time and then evaluate it at t = 1.
Given that r(t) = (4t^2 - t^3)i - 5tj + (t^4 - 2)k, let's differentiate it to find the velocity vector v(t):
v(t) = d(r(t))/dt
= d((4t^2 - t^3)i - 5tj + (t^4 - 2)k)/dt
= (8t - 3t^2)i - 5j + (4t^3)k
Now, to find the momentum at t = 1, we substitute t = 1 into the velocity vector:
v(1) = (8(1) - 3(1)^2)i - 5j + (4(1)^3)k
= 5i - 5j + 4k
(a) The momentum at t = 1 is given by the mass, m, times the velocity vector, v(t):
Momentum = m * v(1)
= 2kg * (5i - 5j + 4k)
= 10i - 10j + 8k
So, the momentum at t = 1 is 10i - 10j + 8k.
To find the force acting on the particle at t = 1, we need to differentiate the velocity vector with respect to time:
a(t) = d(v(t))/dt
= d((8t - 3t^2)i - 5j + (4t^3)k)/dt
= (8 - 6t)i + 0j + (12t^2)k
Now, substituting t = 1 into the acceleration vector, we can find the force:
a(1) = (8 - 6(1))i + 0j + (12(1)^2)k
= 2i + 12k
(b) The force acting on the particle at t = 1 is given by the mass times the acceleration vector:
Force = m * a(1)
= 2kg * (2i + 12k)
= 4i + 24k
So, the force acting on the particle at t = 1 is 4i + 24k.