# Collision of a puck and a brick

• ac7597
In summary, the conversation discusses a problem involving two objects, puck and brick, with given masses and velocities. The total momentum before and after the collision is equal, and the conversation involves finding the angle of the puck's velocity. However, there is a mistake in the calculations and it is suggested to start with the general problem and then substitute in the values. The angle of the puck's velocity is eventually found to be 17.8 degrees.
ac7597
Homework Statement
A hockey puck of mass m=0.36 kg slides in the positive X direction at v=13.8 m/s. It collides with a stationary brick of mass M=1.35 kg.

After the collision, each object slides away to the right. The puck slides at a new speed u=10.410 m/s, at an angle of α above the X-axis, while the brick slides at a speed w=1.340 m/s at a different angle β below the X-axis.

This collision is : inelastic

What is the angle α?
Relevant Equations
momentum=(mass)(velocity)
x(before)y(before)x(after)y(after)
puck (0.36)(13.8)=4.9680(0.36)(10.41)cos(α)(0.36)(10.41)sin(α)
brick00(1.35)(1.34)cos(β)-(1.35)(1.34)sin(β)
total4.968 N*s04.968 N*s0

thus: (0.36)(10.41)sin(α) =(1.35)(1.34)sin(β)
β= sin^(-1)[3.747sin(α)/1.809]= sin^(-1)[2.071sin(α)]

(0.36)(10.41)cos(α) + (1.35)(1.34)cos(β) = 4.968
3.747cos(α) + 1.809cos(sin^(-1)[2.071sin(α)] ) =4.968
3.747cos(α) + 1.809 (1-(2.071sin(α))^(2) )^(1/2) = 4.968
2.071cos(α) + (1-(2.071sin(α))^ (2) )^(1/2) = 2.746
4.29cos^(2)(α) + (1-(2.071sin(α))^(2) ) = 7.54
4.29cos^(2)(α) + 1-4.29sin^2(α) = 7.54
4.29cos^(2)(α) -4.29(1-cos^2(α)) = 6.54
4.29cos^(2)(α) -4.29 + 4.29cos^2(α) = 6.54
8.58cos^(2)(α) = 10.83
α=undefined ?

Here is your mistake. Do you see why this doesn't work?
ac7597 said:
2.071cos(α) + (1-(2.071sin(α))^ (2) )^(1/2) = 2.746
4.29cos^(2)(α) + (1-(2.071sin(α))^(2) ) = 7.54
Suggestion: start with ##m_1,v_1,m_2## and ##v_2## and solve the general problem. Then substitute in the values of the parameters. You are much less likely to make arithmetic errors that way and it makes it easier to understand the equations you are looking at, and possibly avoid errors like the one you have just made.

Last edited:
I squared the equation so [2.071cos(α)]^2 = 4.29cos^2(α) , 2.746^2=7.54 , and [(1-(2.071sin(α))^ (2) )^(1/2)]^2=
(1-(2.071sin(α))^ (2) )

But ##(a+b)^2 \ne a^2+b^2##

if 2.071sin(α)=u
and cos(sin^-1(x))= (1-x^2)^(1/2)
then: (1-u^2)^(1/2)
thus:
2.071cos(α) + (1-u^2)^(1/2) = 2.746
4.29cos^(2)(α) + (1-u^2) = 7.54
4.29cos^(2)(α) + 1-4.29sin^2(α) = 7.54

So you are saying that if ##a+b=c##, then ##a^2+b^2=c^2##.
But given ##1+2=3##
##1^2+2^2=3^2##
##5=9##
So (##a+b=c## implies ##a^2+b^2=c^2##) is not correct. If you square one side of the equation, you have to square the other side, too, like this:
##a+b=c##
##(a+b)^2=c^2##
##a^2+2ab+b^2=c^2##

α=17.8 degrees
thanks

tnich

## 1. What happens when a puck collides with a brick?

When a puck collides with a brick, the force of the impact causes the puck to change direction and possibly speed. The brick will also experience a force, but due to its mass and density, it will likely not move significantly.

## 2. How does the speed of the puck affect the collision with the brick?

The speed of the puck plays a significant role in the collision with the brick. A faster-moving puck will have a greater force and momentum, resulting in a more significant impact and possibly causing more damage to the brick.

## 3. Does the angle of impact between the puck and brick affect the collision?

Yes, the angle of impact can greatly affect the collision between a puck and a brick. A head-on collision will transfer the most force and momentum, while a glancing blow will result in less force being transferred.

## 4. What factors determine the outcome of a collision between a puck and a brick?

Several factors can influence the outcome of a collision between a puck and a brick, including the speed and angle of impact, the mass and density of the objects, and the surface properties of the objects, such as friction and elasticity.

## 5. Can a collision between a puck and a brick be perfectly elastic?

No, it is not possible for a collision between a puck and a brick to be perfectly elastic. Some energy will always be lost in the form of heat, sound, and deformation of the objects involved.

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