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But because particle 2 is initially at rest, this equation becomes

m 1 v 1 x = m 1 v 1 x + m 2 v 2 x .

The components of the velocities along the x size 12{x} {} -axis have the form v cos θ size 12{v`"cos"`θ} {} . Because particle 1 initially moves along the x size 12{x} {} -axis, we find v 1 x = v 1 .

Conservation of momentum along the x size 12{x} {} -axis gives the following equation:

m 1 v 1 = m 1 v 1 cos θ 1 + m 2 v 2 cos θ 2 ,

where θ 1 size 12{θ rSub { size 8{1} } } {} and θ 2 size 12{θ rSub { size 8{2} } } {} are as shown in [link] .

Conservation of momentum along the x size 12{x} {} -axis

m 1 v 1 = m 1 v 1 cos θ 1 + m 2 v 2 cos θ 2

Along the y size 12{y} {} -axis, the equation for conservation of momentum is

p 1 y + p 2 y = p 1 y + p 2 y

or

m 1 v 1 y + m 2 v 2 y = m 1 v 1 y + m 2 v 2 y .

But v 1 y is zero, because particle 1 initially moves along the x size 12{x} {} -axis. Because particle 2 is initially at rest, v 2 y is also zero. The equation for conservation of momentum along the y size 12{y} {} -axis becomes

0 = m 1 v 1 y + m 2 v 2 y .

The components of the velocities along the y size 12{y} {} -axis have the form v sin θ size 12{v`"sin"`θ} {} .

Thus, conservation of momentum along the y size 12{y} {} -axis gives the following equation:

0 = m 1 v 1 sin θ 1 + m 2 v 2 sin θ 2 .

Conservation of momentum along the y size 12{y} {} -axis

0 = m 1 v 1 sin θ 1 + m 2 v 2 sin θ 2

The equations of conservation of momentum along the x size 12{x} {} -axis and y size 12{y} {} -axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level.

Determining the final velocity of an unseen object from the scattering of another object

Suppose the following experiment is performed. A 0.250-kg object m 1 is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg m 2 size 12{ left (m rSub { size 8{2} } right )} {} . The 0.250-kg object emerges from the room at an angle of 45 . size 12{"45" "." 0°} {} with its incoming direction.

The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity ( v 2 and θ 2 ) of the 0.400-kg object after the collision.

Strategy

Momentum is conserved because the surface is frictionless. The coordinate system shown in [link] is one in which m 2 size 12{m rSub { size 8{2} } } {} is originally at rest and the initial velocity is parallel to the x size 12{x} {} -axis, so that conservation of momentum along the x size 12{x} {} - and y size 12{y} {} -axes is applicable.

Everything is known in these equations except v 2 and θ 2 , which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the x - and y -directions.

Solution

Solving m 1 v 1 = m 1 v 1 cos θ 1 + m 2 v 2 cos θ 2 for v 2 cos θ 2 and 0 = m 1 v 1 sin θ 1 + m 2 v 2 sin θ 2 for v 2 sin θ 2 and taking the ratio yields an equation (in which θ 2 is the only unknown quantity. Applying the identity tan θ = sin θ cos θ , we obtain:

tan θ 2 = v 1 sin θ 1 v 1 cos θ 1 v 1 .

Entering known values into the previous equation gives

tan θ 2 = 1 . 50 m/s 0 . 7071 1 . 50 m/s 0 . 7071 2 . 00 m/s = 1 . 129 . size 12{"tan"θ rSub { size 8{2} } = { { left (1 "." "50"" m/s" right ) left (0 "." "7071" right )} over { left (1 "." "50"" m/s" right ) left (0 "." "7071" right ) - 2 "." "00" "m/s"} } = - 1 "." "129"} {}

Thus,

θ 2 = tan 1 1 . 129 = 311 . 312º . size 12{θ rSub { size 8{2} } ="tan" rSup { size 8{ - 1} } left ( - 1 "." "129" right )="311" "." 5° approx "312"°} {}

Angles are defined as positive in the counter clockwise direction, so this angle indicates that m 2 is scattered to the right in [link] , as expected (this angle is in the fourth quadrant). Either equation for the x - or y -axis can now be used to solve for v 2 , but the latter equation is easiest because it has fewer terms.

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Source:  OpenStax, Unit 6 - momentum. OpenStax CNX. Jan 22, 2016 Download for free at https://legacy.cnx.org/content/col11961/1.1
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