Question

(II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball? (b) What fraction of the original kinetic energy ( KE/KE) gets transferred to the second ball?

Answer 1

Answer:

A. 0.84 kg

B. 75 %

Explanation:

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Answer 2

The mass of the second ball initially at rest before the collision is 0.84 kg.

The fraction of the original kinetic energy transferred to the second ball is 0.75.

The given parameters;

• mass of the ball, m = 0.28 kg
• mass of the second ball, = m
• initial velocity of the first ball, = u

Apply the principle of conservation of linear momentum;

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\0.28u + m(0) = 0.28v_1 + m(\frac{u}{2} )\\\\0.28 u = 0.28v_1 + 0.5mu$

apply one-dimensional velocity;

$u_1 + v_1 = u_2 + v_2\\\\u + v_1 = 0 + 0.5u\\\\v_1 = 0.5u - u\\\\v_1 = -0.5u$

From the first equation, substitute v₁;

$0.28u = 0.28(-0.5u) + 0.5mu\\\\0.28u = -0.14u + 0.5mu\\\\0.28u + 0.14u = 0.5mu\\\\0.42u = 0.5mu\\\\m = \frac{0.42u}{0.5u} \\\\m = 0.84 \ kg$

The original kinetic energy of the ball is calculated as follows;

$K.E_i = \frac{1}{2} mu^2\\\\K.E _i = 0.5 (0.28)u^2\\\\K.E _i = 0.14u^2$

The kinetic energy transferred to the second ball is calculated as follows;

$K.E_f = \frac{1}{2} mu^2\\\\K.E_f = \frac{1}{2} m(\frac{u}{2} )^2\\\\K.E_f =\frac{1}{2} \times 0.84\times \frac{u^2}{4} \\\\K.E_f = 0.105u^2$

The fraction of the original kinetic energy transferred to the second ball is calculated as follows;

$= \frac{KE_f}{K.E_i} = \frac{0.105u^2}{0.14u^2} = 0.75 = \frac{3}{4}$

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