Billiard ball A of mass $m_A = 0.120 \textrm{ kg}$ moving with speed $v_A = 2.80 \textrm{ m/s}$ strikes ball B, initially at rest, of mass $m_B = 0.140 \textrm{ kg}$. As a result of the collision, ball A is deflected off at an angle of $30.0 ^\circ$ with a speed ${v^{'}_{A}} = 2.10 \textrm{ m/s}$.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Here are the equation's showing conservation of momentum in the x-direction and in the y-direction. The x and y directions are independent and so momentum is conserved separately in each of these directions. So we have the incoming mass A colliding with mass B which is initially at rest and ball A is deflected up at this angle, Θ A above the horizontal, and has some speed v A prime after the collision and mass B is moved down at this angle, Θ B below horizontal, with some speed v B prime. So in the x-direction, we have both of these blocks having an x-component of their momenta and we have m A times v A prime times the component of this momentum in the x-direction and I guess I could write a triangle here to show the components; this is the component of momentum of block A in the y-direction and that equals m Av A prime times sin Θ A and then in the x-direction for A, we have momentum of block A in the x-direction is m A times v A prime times cos Θ A and that's what we have here. And it's the same idea for ball B; it's m B times v B prime times cos Θ B. And these two x-momenta combined after the collision, have to equal the total x-momentum that this sytem has before the collision and that is entirely in just mass A because mass B has no velocity and therefore no momentum before the collision. So before the collision, on the left hand side of the equal sign, there's only gonna be the component of the mass A's momentum in the x-direction and that's gonna be m Av A; no trigonometry needed because it's entirely along the x-axis and that's how we define the x-axis is to be in the direction of velocity before the collision so that this left hand side is nice and convenient with no trigonometry there. For the y-direction, there's no momentum in the y-direction before the collision so that means whatever momentum this block A has upwards has to equal the momentum of the ball B going downwards so that they cancel each other out and there we go. Now part (b) of this question asks us to figure out what is this speed of ball B and what is its angle? So based on this equation, we can solve for v B prime by dividing both sides by m Bsin Θ B and we get v B prime is m Av A prime times sin Θ A divided by m Bsin Θ B and that is useful because we can substitute it into the x-direction formula and then figure out Θ B first and then we'll figure out v B prime afterwards. So in red here, I have shown the substitution for v B prime So I have copied this x-direction momentum conservation formula but instead of writing v B prime, I'm writing m Av A prime sin Θ A over m B sin Θ B in its place. So this looks big and messy but that's the way it has to be done. And a trigonometric identity which will clean things up a slight bit is to know that cos of an angle divided by sin of the same angle is 1 over tan of that angle. So I'll rewrite this equation on this line with that substitution— 1 over tan Θ B instead of cos Θ B over sin Θ B and also switching the sides around so this term is on the left side but before we do the switching, we are gonna move this term to the left side by subtracting it from both sides so that makes it minus m Av A prime cos Θ A and then we'll switch the sides around. So next step would be to multiply both sides by tan Θ B and divide both sides by this whole thing so that we get that tan Θ B is m Av A prime sin Θ A over m Av A minus m Av A prime cos Θ A and then we plug in some numbers. And we get that Θ B is inverse tangent of the 0.120 kilograms—mass of ball A— times its speed after the collision 2.10 meters per second times sin of its angle of deflection 30.0 degrees and to divide that by 0.120 kilograms times speed of ball A before the collision of 2.80 meters per second minus 0.120 kilograms times the speed after the collision 2.10 meters per second times cos of the angle of deflection and that gives 46.9 degrees must be the angle of deflection for ball B and that's below the horizontal. And for the speed of ball B, we have a formula for that here and it's gonna be the mass of ball A, 0.120 kilograms, times the speed of ball A after deflection, 2.10 meters per second, times sin 30.0 degrees divided by the mass of ball B, 0.140 kilograms, times sin of its angle that we just figured out and I'm writing a whole bunch of extra digits in here 46.936 instead of just 46.9 to avoid intermediate rounding error and in fact in my calculator, I just do second function negative sign that gives us the blue "ANS", which stands for "answer" of the previous line so I'm plugging in all these digits here actually so it's 46.93568657 which gets plugged into the sin there. So this gives us 1.23 meters per second. So velocity of ball B after the collision is 1.23 meters per second, 46.9 degrees below the horizontal.