A hockey puck of mass m1, moving with initial speed u, collides with another puck of mass m2, initially at rest. After the collision, the first puck moves with known speed v in the direction perpendicular to the initial. Whats the final speed of the second puck?
Well, isn't that a slippery situation! Let's break it down. Since the collision is described as occurring in a perpendicular direction, we can treat it as a 2-dimensional problem.
Now, we know that the first puck, with mass m1, is initially moving with speed u. After the collision, it moves with speed v in the perpendicular direction. But what about the second puck, with mass m2, initially at rest? Ah, poor guy, just minding its own business.
To determine the final speed of the second puck, we can use the law of conservation of momentum. This states that the total momentum before the collision should be equal to the total momentum after the collision.
Now, since the second puck is initially at rest, its initial momentum is zero. So, we can say that:
m1 * u = m1 * v + m2 * vf2
Here, vf2 represents the final speed of the second puck. Now, if we rearrange the equation, we can solve for vf2:
m1 * u - m1 * v = m2 * vf2
vf2 = (m1 * u - m1 * v) / m2
And there you have it! The final speed of the second puck can be determined using this equation. Just plug in the values for m1, m2, u, and v, and calculate away. Just be careful not to get tangled on the ice while doing the math!
To find the final speed of the second puck, we can use the principle of conservation of linear momentum.
The linear momentum before the collision is equal to the linear momentum after the collision. This can be expressed as:
m1 * u = m1 * v1 + m2 * v2
where:
m1 is the mass of the first puck
u is the initial speed of the first puck
v1 is the final velocity of the first puck
m2 is the mass of the second puck
v2 is the final velocity of the second puck
Since the second puck is initially at rest, we can substitute the value of v1 with v, the speed of the first puck after the collision. This gives us:
m1 * u = m1 * v + m2 * v2
Now, we need additional information to solve for v2.
To find the final speed of the second puck, we can use the principles of conservation of momentum and conservation of kinetic energy.
The principle of conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum after the collision, assuming there are no external forces acting on the system.
Before the collision, the momentum of the first puck is given by p1 = m1 * u (where p is the momentum and m is the mass).
Since the second puck is initially at rest, its momentum is zero, p2 = 0.
After the collision, the first puck moves with speed v in the perpendicular direction. The momentum of the first puck is given by p1' = m1 * v.
The final speed of the second puck, denoted as v2', can be found using momentum conservation:
p1 + p2 = p1' + p2'
m1 * u + 0 = m1 * v + m2 * v2'
Simplifying the equation, we have:
m1 * u = m1 * v + m2 * v2'
Now, let's consider the conservation of kinetic energy. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
The initial kinetic energy of the first puck is given by:
KE1 = (1/2) * m1 * u^2
The initial kinetic energy of the second puck is zero since it is at rest.
After the collision, the kinetic energy of the first puck is given by:
KE1' = (1/2) * m1 * v^2
The kinetic energy of the second puck can be calculated as:
KE2' = (1/2) * m2 * v2'^2
Using conservation of kinetic energy, we have:
KE1 + KE2 = KE1' + KE2'
(1/2) * m1 * u^2 + 0 = (1/2) * m1 * v^2 + (1/2) * m2 * v2'^2
Simplifying the equation, we have:
m1 * u^2 = m1 * v^2 + m2 * v2'^2
Now we have a system of two equations:
m1 * u = m1 * v + m2 * v2'
m1 * u^2 = m1 * v^2 + m2 * v2'^2
We can solve these two equations simultaneously to find the final speed of the second puck, v2'. The equation involves some algebraic manipulation to isolate v2' and solve for it explicitly.
unlikely but x momentum before = m1 u1
y momentum before = 0
x momentum after = m2 u2b so u2 b = (m1/m2)u1
y momentum after = m1 v +m2 v2b = 0
so
v2b = - (m1/m2)v so V2b is known if v is known