# Solving Hibbeler 15-79 Collision Problem w/ Mass 3m & e

• Bebalo
In summary: The block has a mass of 3m and is resting on a smooth surface. The goal is to determine the block's velocity just after the collision, with a given coefficient of restitution "e". The block is a 45-45-90 triangle and the sphere strikes the middle of the hypotenuse. The surface the block is resting on is perpendicular to the motion of the falling sphere. The solution given in the book is: velocity of block = ((1-e)/7)*initial v of sphere. The approach involves using conservation of linear momentum and considering the angle of collision in the calculation.
Bebalo
Hi, I'm having trouble with an assignment problem. If you happen to have Hibbeler's Tenth Edition of "Dynamics" it is problem 15-79. For those who don't have a copy of the book, the problem is:

The sphere of mass "m" falls and strikes the triangular block with a vertical velocity v. If the block rests on a smooth surface and has a mass 3m, determine its velocity just after the collision. The coefficient of restitution is e.

The block is a 45-45-90 triangle and the sphere is falling and striking the middle of the hypotenuse. The surface the block is resting on is perpendicular to the motion of the falling sphere (ie: surface is like counter on which block is sitting).

I've tried using conservation of momentum in the normal direction (through the center of mass of the triangle and sphere, makes 45 degree angle with surface) but this results in the final velocity of the block being "into" the surface. How do I deal with this.

The solution given in the book is: velocity of block = ((1-e)/7)*initial v of sphere.

Thanks in advace for the help.

Bebalo said:
The solution given in the book is: velocity of block = ((1-e)/7)*initial v of sphere.
Tough problem. Made all the more difficult due to my lack of familiarity working with coefficient of restitution especially at oblique angles.

I assume the sphere/ball deflects at a 45 degree angle to the surface so its motion immediately after the collision is horizontal. Its speed after a headon collision would be given by the coefficient of restitution (e = speed of approach/speed of separation) but what is the effect of the angle? If you can get that, I think you can figure it out using conservation of linear momentum.

It looks to me like in a head on collision, $KE_{ball}+KE_{\Delta} = e^2KE_{initial}$. In a 45 degree collision, the loss of energy would be 1/2 (.707^2) of that of a head on collision. But that is just a guess. And I am too beat right now to wrap my head around it.

AM

Hello,

I can understand the difficulty you are facing with this problem. It is important to approach this problem systematically and use the principles of physics to solve it.

Firstly, we need to consider the conservation of momentum in both the x and y directions. In the x direction, we have the sphere with mass m and initial velocity v and the block with mass 3m and initial velocity of 0. After the collision, the sphere and the block will have a common velocity in the x direction, let's call it Vx.

Using the conservation of momentum equation, we can write:

mv = (m+3m)Vx

Simplifying this, we get:

Vx = v/4

Now, in the y direction, the only force acting is the weight of the sphere which is balanced by the normal reaction force from the surface. This means there is no net force in the y direction and hence the velocity in the y direction will remain constant.

Therefore, we can write:

Vy = v

Now, we can use the coefficient of restitution (e) to relate the velocities before and after the collision. The coefficient of restitution is defined as the ratio of relative velocity after collision to the relative velocity before collision. In this case, the relative velocity is the difference between the velocities of the sphere and the block.

So, we can write:

e = (Vx - Vy)/(v - 0)

Solving for Vx, we get:

Vx = (1-e)v

Substituting the value of v/4 from the previous equation, we get:

(1-e)v = v/4

Solving for e, we get:

e = 3/4

Now, we can use this value of e in the given solution to calculate the final velocity of the block:

Velocity of block = ((1-e)/7)*initial velocity of sphere

= ((1-3/4)/7)*v

= (1/28)*v

Therefore, the final velocity of the block after the collision is (1/28)*v.

I hope this explanation helps you understand the solution to this problem. Remember to always approach problems like this systematically and use the principles of physics to guide you. Good luck with your assignment!

## 1. How do you approach solving a collision problem involving mass and coefficient of restitution?

The first step in solving a collision problem involving mass and coefficient of restitution is to draw a diagram of the system and label all known and unknown variables. Then, use conservation of momentum and conservation of energy equations to set up a system of equations. Finally, solve for the unknown variables using algebraic manipulation.

## 2. What is the role of the coefficient of restitution in a collision problem?

The coefficient of restitution, denoted by e, is a measure of the elasticity of a collision. It indicates the ratio of the final relative velocity of two objects after a collision to the initial relative velocity. In other words, it shows how much kinetic energy is conserved in a collision. This information is crucial in solving a collision problem involving mass and velocity.

## 3. How is momentum conserved in a collision problem?

Momentum, denoted by p, is a vector quantity that is conserved in a closed system. This means that the total momentum of a system before a collision is equal to the total momentum after a collision. In mathematical terms, this can be expressed as:

Initial momentum = Final momentum
(mass1 * velocity1i) + (mass2 * velocity2i) = (mass1 * velocity1f) + (mass2 * velocity2f)

## 4. Can the collision problem be solved without knowing the coefficient of restitution?

Yes, the collision problem can still be solved without knowing the coefficient of restitution. In this case, the value of e can be treated as an unknown variable and can be solved for using the given information and other equations, such as the conservation of energy equation. However, having the value of e can make the problem easier to solve and can provide additional information about the collision.

## 5. What are some common mistakes to avoid when solving a collision problem with mass and coefficient of restitution?

Some common mistakes to avoid when solving a collision problem with mass and coefficient of restitution include using the incorrect signs for velocities, not considering the direction of motion in the momentum equations, and assuming that the objects have the same mass. It is also important to double check units and make sure they are consistent throughout the problem. It is always a good idea to draw a diagram and label all variables to avoid any confusion.

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