Ball A, of mass 1.5 kg, moves along the positive x-axis with an initial speed of 2.0 m/s. It collides with Ball B, of mass 0.80 kg, which is initially at rest. After they collide, Ball B moves off at an angle of 73° with respect to the positive x-axis, with a speed of 1.5 m/s.

a) What is the final velocity of Ball A?

b) Is the collision elastic?

a) For the collision of the two masses, we use momentum conservation (there are no net external forces on the system of the two balls):

m_Av + 0 = m_Av_A + m_Bv_B

where

v = 2.0 m/s i

m_A = 1.5 kg

m_B = 0.8 kg

v_B = 1.5 m/s   at 73°

Let's work out the x- and y-components of v_B:

v_Bx = v_Bcos(73°) = (1.5 m/s)cos(73°) = 0.438 m/s

v_By = v_Bsin(73°) = (1.5 m/s)sin(73°) = 1.434 m/s

so the x-component of our conservation of momentum equation is:

m_Av = m_Av_Ax + m_Bv_Bx

or

(1.5 kg)(2.0 m/s) = (1.5 kg)v_Ax + (0.8)(0.438 m/s)

solving for v_Ax gives

v_Ax = 1.77 m/s

The y-component of our conservation of momentum equation is:

0 = m_Av_Ay + m_Bv_By

or

0 = (1.5 kg)v_Ay + (0.8 kg)(1.434 m/s)

thus

v_Ay = - 0.765 m/s

Thus

v_A = 1.77 m/s i - 0.765 m/s j

We will need the magnitude of this for part b), so let's calculate it:

v_A = [( 1.77 m/s)² + (-0.765 m/s)²)^1/2 = 1.93 m/s

b) To test whether the collision is elastic, we find the kinetic energy before and after:

K_i = ½m_Av² = ½(1.5 kg)(2.0 m/s)² = 3.00 J

K_f = ½m_Av_A² + ½m_Bv_B²

= ½(1.5 kg)(1.93 m/s)² + ½(0.8 kg)(1.5 m/s)²

= 3.69 J

These are not the same, so energy is not conserved, the collision is not elastic.

Next Problem
Test 2
Physics 101 Home page
Physics Department Home Page