Question

A ball is projected at an immovable wall with a speed vi and bounces back the wall in such a manner that it only has 1/3 of its original linear momentum. a) Determine what fraction of the kinetic energy is lost in the collision.

Answer 1

The fraction of the kinetic energy of the ball lost during the collision is $\frac{1}{3}$.

The given parameters;

• initial speed of the ball, = vi
• final momentum of the ball, Pf = ¹/₃Pi

The initial and final momentum of the ball is calculated as;

$P_i = m_ivi$

$P_f = m_fv_f = \frac{1}{3} m_iv_i$

The initial and final kinetic energy of the ball is calculated as;

$K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i$

The change in the kinetic energy is calculated as;

$\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i$

Thus, the fraction of the kinetic energy of the ball lost during the collision is $\frac{1}{3}$.

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