# A body of mass 2kg is moving along north-east direction with a speed of √2 m/s.plz read below....A body of mass 2kg is moving along north-east direction with a speed of √2 m/s. a force of 0.2N...

A body of mass 2kg is moving along north-east direction with a speed of √2 m/s.plz read below....

A body of mass 2kg is moving along north-east direction with a speed of √2 m/s. a force of 0.2N is applid one the body due west for 10 sec. find the final velocity of the body and is direction.

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Apply the impulse-momentum change theorem. The formula is:

`sumF * t =m (v_2 - v_1)`

The given in the problem are:

m=2 kg initial velocity `v_1 = sqrt2` (north east direction)

t =10 sec F = 0.2N (due west)

Take note that we can not directly substitute the given values to the formula. The applied force is not in the same direction with the motion of the body. So to solve for the final velocity `v_2` , its corresponding x and y components must be determined first.

> Along the x-axis,

`F_x *t = m(v_(2x) - v_(1x))`

`Fcos theta *t = m(v_(2x) - v_1cos phi)`

Since the direction of the force is due west, `theta` is `180^o` . While `phi` is `45^o`.

`0.2cos 180 * 10 = 2 (v_(2x) - sqrt2cos45)`

`-2 = 2(v_(2x)-1)`

To simplify, divide both sides by 2. Then, solve for `v_(2x)` .

`-1 = v_(2x) -1`

`-1+1 = v_(2x)`

`0 = v_(2x)`

The x-component of `v_2` is `v_(2x)=0` m/s .

> Along the y-axis,

`F_y*t = m(v_(2y) - v_(1x))`

`Fsin theta * t = m (v_(2y) - v_1sin phi)`

`0.2sin180*10= 2 (v_(2y) - sqrt2sin45)`

`0 = 2 (v_(2y)-1)`

`0 = v_(2y) - 1`

`1 = v_(2y)`

The y-component of the final velocity is `v_(2y)=1` m/s .

Next, solve for the resultant of the final velocity.

`v_2 = sqrt(v_(2x)^2 + v_(2y)^2 )= sqrt(0^2+1^2) = sqrt(1)`

`v_2 = 1`

To solve for the direction,

`v_(2y) = v_2sin alpha`

`alpha = sin^(-1) v_(2y)/v_2 = sin^(-1) 1/1 = sin^(-1)1 = 90^o`

**Hence, the final velocity of the body is 1m/s and travelling due north.**

Using Newton's second law,

`lt- F= ma`

0.2 = 2*a

a = 0.1 m/s^2

The body has a velocity `sqrt2` at north east direction.

The component velocity to the east = `sqrt2*cos45` = 1 m/s

Lets assume the final velocity of the body is v at west.

By velocity equation v= u +at to the west direction

v = -1+ 0.1*10

= 0 m/s

Lets assume the final velocity of the body is v at North

`uarr F= ma`

0 = 2*a

a = 0 m/s^2

`uarr v= u+at`

v = sqrt2 cos45

= 1m/s

**So the resultant velocity = sqrt(1^2+0^2) = 1m/s**

**Direction of the velocity = North direction**

Assumption

There is no frictional forces acting on the body.