Question 8.5: Collision in a horizontal plane This is a two-dimensional ve...
Collision in a horizontal plane
This is a two-dimensional version of a conservation of-momentum problem. Now we will have to keep track of both the x and y components of the momentum of each object in the system. Figure 8.7 shows two chunks of ice sliding on the surface of a frictionless frozen pond. Chunk A, with mass m_A = 5.0 kg, moves with initial velocity \upsilon_{A,i} = 2.0 m/s parallel to the x axis. It collides with chunk B, which has mass m_B = 3.0 kg and is initially at rest. After the collision, the velocity of A is found to be \upsilon_{A,f} = 1.0 m/s in a direction at an angle α = 30° with the initial direction. What is the final velocity of B (magnitude and direction)?
SET UP The velocities are not all along a single line, so we have to use both x and y components of momentum for each chunk of ice. There are no horizontal external forces, so the total horizontal momentum of the system is the same before and after the collision. Conservation of momentum requires that the sum of the x components before the collision must equal their sum after the collision, and similarly for the y components. We must write a separate equation for each component. We draw the coordinate axes as shown in Figure 8.7. We also express the final velocities in terms of their components.
SOLVE We start by writing expressions for the total x component of momentum before and after the collision. For the x components, we have
m_A(\upsilon _{A,i,x})+mB(\upsilon _{B,i,x})
=(5.0 kg)(2.0 m/s)+(3.0 kg)(0) (before),
m_A(\upsilon _{A,f,x})+m_B(\upsilon _{B,f,x})
=(5.0 kg)(1.0 m/s)(\cos 30°)+(3.0 kg)(\upsilon _{B,f,x}) (after),
Equating these two expressions and solving for \upsilon _{B,f,x}, we find that
\upsilon _{B,f,x} = 1.89 m/s (x component of final velocity of B)
Conservation of the y component of total momentum gives
m_A(\upsilon _{A,i,y})+m_B(\upsilon _{B,i,y})
=(5.0 kg)(0) + (3.0 kg)(0) (before),
m_A(\upsilon _{A,f,y})+m_B(\upsilon _{B,f,y})
=(5.0 kg)(1.0 m/s)(\sin 30°)+(3.0 kg)(\upsilon _{B,f,y}) (after),
Equating these two expressions and solving for \upsilon _{B,f,y}, we obtain
\upsilon _{B,f,y} = -0.83 m/s (y component of final velocity of B)
We now have the x and y components of the final velocity \overrightarrow{\pmb{\upsilon }}_{B,f} of chunk B. The magnitude of \overrightarrow{\pmb{\upsilon }}_{B,f} is
\left|\overrightarrow{\pmb{\upsilon }}_{B,f}\right| = \sqrt{(1.89 m/s)^2+(-0.83 m/s)^2}
= 2.1 m/s (final speed of B)
and the angle β of its direction from the positive x axis is
\beta =\tan^{-1}\frac{-0.83 m/s}{1.89 m/s}=-24° .
REFLECT It’s essential to treat velocity and momentum as vector quantities, representing them in terms of components. You can then use these components to find the magnitudes and directions of unknown vector quantities.
Practice Problem: If chunk B has an initial velocity of magnitude 2.0 m/s in the +y direction instead of being initially at rest, find its final velocity (magnitude and direction). Answers: 2.2 m/s, 32°.