To solve this problem, we must remember about the law of
conservation of momentum. The initial momentum mist be equal to the final
momentum, that is:
m1 v1 + m2 v2 = (m1 + m2) v’
where v’ is the speed of impact
Since we are not given the masses of each car m1 and m2,
so let us assume that they are equal, such that:
m1 = m2 = m
Which makes the equation:
m v1 + m v2 = (2 m) v’
Cancelling m and substituting the v values:
50 + 48 = 2 v’
2 v’ = 98
v ‘ = 49 km/h
<span>The speed of impact is 49 km/h.</span>
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass 
kg, charge +e = 
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 
m/s. The proton comes momentarily to rest at a distance 
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 
m apart?
Explanation:
The given data is as follows.
Mass of proton = 
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
= 
where, 
U = 
Putting the given values into the above formula as follows.
U =
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
Answer:
To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.
Explanation:
Given ,
F = 130N
M = 24kg
A = ?
F = m× a
then ,
130N = 24kg ×a
a = 130/24 = 5 m/s.
Answer:
only the weight of the ball will act on the ball
Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown
