Collision of a puck and a brick

[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)=4.968	0	(0.36)(10.41)cos(α)	(0.36)(10.41)sin(α)
brick	0	0	(1.35)(1.34)cos(β)	-(1.35)(1.34)sin(β)
total	4.968 N*s	0	4.968 N*s	0

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 ?

 

[tnich]

Here is your mistake. Do you see why this doesn't work?

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.

 

[ac7597]

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) )

 

[tnich]

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

 

[ac7597]

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

 

[tnich]

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:
$a+b=c$
$(a+b)^2=c^2$
$a^2+2ab+b^2=c^2$

 

[ac7597]

α=17.8 degrees
thanks