Question

Two pucks collide on an air hockey table. Puck A has a mass of 28 g and is initially traveling in the +x direction at 7.8 m/s. Puck B has a mass of 112 g and is initially at rest. After the pucks collide, puck A moves away at an angle of 45$^\circ$ above the +x axis, while puck B travels at an angle of 42$^\circ$ below the +x axis.
Calculate puck A's final speed and puck B's final speed. Also, what fraction of kinetic energy was lost in this collision?

Answer 1

The mass of puck A is m_A = 28 m = 0.028 kg

The mass of puck B is m_B = 112 g = 0.112 kg

The initial speed of puck A is u_A = 7.8 m/s (along positive X axis)

The initial speed of peck B is zero

The angle made by puck A with positive X axis after the collision is θ_A = 45^o

The angle made by puck B with positive X axis after the collision is θ_B = -42^o

Let v_A and v_B are the velocities of puck A and puck B respectively after the collision

Apply the law of conservation of linear momentum in Y axis

m_A(0) + m_B(0) = m_Av_Asinθ_A + m_Bv_Bsinθ_B

m_Av_Asinθ_A + m_Bv_Bsinθ_B = 0

(0.028 kg)v_Asin(45^o) + (0.112 kg)v_Bsin(-42^o) = 0

0.0198v_A - 0.0749v_B = 0

v_A = 3.7828v_B

Apply the law of conservation of linear momentum in X axis

m_Au_A + m_B(0) = m_Av_Acosθ_A + m_Bv_Bcosθ_B  

m_Au_A = m_Av_Acosθ_A + m_Bv_Bcosθ_B

(0.028 kg)(7.8 m/s) = (0.028 kg)v_Acos(45^o) + (0.112 kg)v_Bcos(-42^o)  

0.2184 = 0.0198v_A + 0.0832v_B

0.2184 = 0.0198(3.7828v_B) + 0.0832v_B

v_B = 1.38 m/s

Therefore

v_A = 3.7828(1.38 m/s)

v_A = 5.22 m/s

The initial kinetic energy of the system is K_i = 0.5m_Au_A²

K_i = 0.8518 J

The final kinetic energy of the system is K_f = 0.5m_Av_A² + 0.5m_Bv_B²

K_f = 0.4881 J

The loss in the kinetic energy is ΔK = 0.8518 J - 0.4881 J = 0.3637 J

The fraction of the loss in kinetic energy is

K_fraction = ΔK/K_i

K_fraction = 0.3637/0.8518

K_fraction = 0.427