Question

a toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as xy plane. the 4 kg pick has a velocity of 3.00 i m/s at one instant. eight seconds later its velocity is (8i + 10j) m/s. assuming the rocket engine exerts a constant horizontal force find the components of the force and its magnitude

Answer 1

@cshalvey

Answer 2

I got F = 5/2 i + 5 j..........I'm not sure if I'm right.

Answer 3

@mukushla

Answer 4

we have ------- V(0) = 3i and V(8) = 8i + 10j

F = constant ---> F = ma ---> F = mdv/dt ---> Fdt = mdv

integration and then we have :
F*t = mv + c
F*0 = 4*(3i) + c (1)
F*8 = 4*(8i + 10j) + c (2)

solve for c from equation 1 and put in the equation 2

8F=20i+40j
F=5/2i+5j

u r right @Vaidehi09

Answer 5

uhmm what happened here? will someone be kind enough to explain the steps? haha sorry...like i said im new to this

Answer 6

Given: acceleration = constant.
Let it be \(a = x\hat{i}+y\hat{j}\)
We know that for constant acceleration, \(\vec{v} = \vec{u} + \vec{a}t\) where \(\vec{v}\) is the final velocity, \(\vec{u}\) is the initial velocity and t = time taken.
Given: \(\vec{u} = 3\hat{i}, \vec{v} = 8\hat{i}+10\hat{j}, t = 8\)

Substituting, we get: \(\vec{a} = (5\hat{i}+10\hat{j})/8\)

Now, force \(\vec{F} = m\vec{a}\) and m = 4kg

So, \(\vec{F} = (5/2)\hat{i} + 5\hat{j}\)

\(|\vec{F}| = \sqrt{(5/2)^{2}+5^{2}}= 5\sqrt{5}/2\)