# Explain how conservation of momentum applies to a handball bouncing off a wall.

Kevin Nash
7 years ago
Law of conservation of linear momentum states that, in an isolated system (no external force), the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other.
The momentum of particle (p) is equal to the mass of particle (m) times the velocity of particle (v).
So, p = mv …… (1)
Let us consider m is the mass of the ball and v is the velocity of the ball when the ball is collides with wall.
So using equation (1), the momentum of the ball before collision (p1) will be,
p1= mv …… (2)
After collision, when the ball re bounces, the velocity of the ball will be, -v.
So again using equation (1), the momentum of the ball after collision (p2) will be,
p2= -mv …… (3)
Conservation of linear momentum states that, the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other.
p1 + p2 = 0
So, mv + (-mv) = 0 …… (4)
From equation (4) we observed that, linear momentum of the hand ball is conserved.
Rizwan Mahsood
41 Points
6 years ago
The rebouncing of a handball from the wall is a type of elastic collision in which both momentum and kinetic energy remains constant.For conservation of momentum isolated system must be required.We know the conservation expression for a single particle,P=mv (before collision)P=-mv (after collision)The negative sign is because of velocity is a vector quantity and it is opposite to the initial velocity vector.Thus the net momentum becomes,Pnet=(mv)+(-mv)=0
Rizwan Mahsood
41 Points
6 years ago
And also we can write this expression asP=mv(before collision) -mv(after collision)If P=0 (conservation of momentum)Then we can write,mv(before collision)-mv(after collision)=0Or,mv(before collision)=mv(after collision)Hence it means momentum is conserved