Question

A 2 kg ball going 10 m/s follows a path that is perpendicular to the surface of a wall. It impacts the wall and looses 20 % of its momentum in the collision. What is the balls impulse?

Answer 1

The impulse delivered by the wall to the ball can be calculated using the formula Impulse = Δp = m * Δv, where Δp is the change in momentum, m is the mass of the ball, and Δv is the change in velocity. In this case, the ball loses 20% of its momentum, resulting in a change in velocity of 2 m/s. Therefore, the impulse delivered by the wall to the ball is 4 kg·m/s.

Step-by-step explanation:

The impulse delivered by the wall can be calculated using the formula:

Impulse = Δp = m * Δv

Where Δp is the change in momentum, m is the mass of the ball, and Δv is the change in velocity.

In this case, the ball loses 20% of its momentum, which means the change in velocity is 20% of the initial velocity. Since the ball was initially moving perpendicular to the wall, the change in velocity is purely in the x-direction.

Therefore, the impulse delivered by the wall is:

1. Calculate the change in velocity:
   Δv = 0.2 * 10 m/s = 2 m/s
2. Calculate the impulse:
   Impulse = 2 kg * 2 m/s = 4 kg·m/s

Answer 2

If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).

For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).

In other words, the total momentum in the x direction will be the same before and after the collision.

\Large \Sigma p_{xi}=\Sigma p_{xf}Σp

xi

​

=Σp

xf

​

\Sigma, p, start subscript, x, i, end subscript, equals, \Sigma, p, start subscript, x, f, end subscript

Also, the total momentum in the y direction will be the same before and after the collision.

\Large \Sigma p_{yi}=\Sigma p_{yf}Σp

yi

​

=Σp

yf

​

\Sigma, p, start subscript, y, i, end subscript, equals, \Sigma, p, start subscript, y, f, end subscript

In solving 2 dimensional collision problems, a good approach usually follows a general procedure:

Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.

Write down all the values you know and decide exactly what you need to find out to solve the problem.

Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.

Step-by-step explanation:

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