Answer:
1) Newton's first law of motion states an object will remain at rest or in uniform will be in uniform motion in a straight line unless a force acts on it
2) Newton's second law states the acceleration of an object is directly proportional to the applied force acting on an object and inversely proportional to the mass of the object
Explanation:
1) With Newton's first law, we are able arrange things within a space and schedule meetings in time knowing that they will remain in place unless an external force changes their positions
2) An example of Newton's second law of motion is that small objects such as a ball are easily accelerated and can be given appreciable acceleration for flight by single, one time contact (such as kicking the ball) while larger objects such as a rock require sustained force application to change their location.
Answer:
The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg.
As an object falls in response to gravitational force, its velocity increases
and its acceleration is unchanged. On Earth, its acceleration is 9.8 m/s².
Answer:
In the analytical method,
- Resolve the vectors into the perpendicular components of the Cartesian coordinates.
- Calculate the magnitude of the resultant vector using the Pythagoras theorem.
Explanation:
- There are two methods to find the magnitude of the resultant vector.
- One is the geometrical method and the other one is the analytical method.
- In the geometrical method, all the vectors are connected the head to tail with the appropriate magnitude and the resultant vector is obtained by joining the initial point and the final point by a vector in the reverse direction. The magnitude of the resultant vector is given by the length of the line.
- In the analytical method, all the vectors are resolved into the perpendicular components.
- Using Pythagoras theorem, the magnitude of the resultant vector can be obtained
- If A and B are the two vectors forming an angle ∅ between them, then the magnitude of the resultant vector is given by the formula

a)
We use the formula :
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the values in:
4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f
Calculating this we get:
32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f
Rearrange for v2f:
v2f = 
This gives us 8.0 m/s as the final velocity of the second ball.
b)
Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.
This means we use the formula:
Ek =
+
=
+ 
Substituting in values:
Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2
This simplifies to:
Ek= 128J + 0J = 0J + 128J
This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.