After the collision the magnitude of the momentum of the system is Mv
Given:
mass of 1st object = M
speed of 1st object = v
mass of 2nd object = M
speed of 2nd object = 0
To Find:
magnitude of the momentum after collision
Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
Applying conservation of linear momentum
Mv + M(0) = 2MV
Mv = 2MV
V = v/2
So, after collision momentum is
p = 2MV = 2xMxv/2 = Mv
So, after collision momentum is Mv
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Answer:270joules
Explanation:KE = 1/2mv^2
KE = 1/2(60kg)(3.0)^2 = 270 Joules
<span>There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)</span>
g Generally the accepted value of acceleration due to gravity is 9.801 
as per the question the acceleration due to gravity is found to be 9.42
in an experiment performed.
the difference between the ideal and observed value is 0.381.
hence the error is -
=3.88735 percent
the error is not so high,so it can be accepted.
now we have to know why this occurs-the equation of time period of the simple pendulum is give as-![T=2\pi\sqrt[2]{l/g}](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Csqrt%5B2%5D%7Bl%2Fg%7D)

As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.
if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801 
Answer:
Explanation:
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