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(B) Given: Mass $m = 1\, kg$,initial velocity $v_1 = 8\, m/s$,compression $x = 3\, m$.Initial momentum $P_1 = m v_1 = 1 	imes 8 = 8\, kg\cdot m/s$.Fi

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$16/3\, N/m$

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(B) Given: Mass $m = 1\, kg$,initial velocity $v_1 = 8\, m/s$,compression $x = 3\, m$.Initial momentum $P_1 = m v_1 = 1 	imes 8 = 8\, kg\cdot m/s$.Final momentum $P_2 = P_1 / 2 = 4\, kg\cdot m/s$.Since $P_2 = m v_2$,we have $v_2 = P_2 / m = 4 / 1 = 4\, m/s$.By the law of conservation of energy,the loss in kinetic energy is equal to the potential energy stored in the spring:$\frac{1}{2} m v_1^2 - \frac{1}{2} m v_2^2 = \frac{1}{2} k x^2$$m(v_1^2 - v_2^2) = k x^2$$1 	imes (8^2 - 4^2) = k 	imes 3^2$$1 	imes (64 - 16) = 9k$$48 = 9k$$k = 48 / 9 = 16 / 3\, N/m$.

$A$ $1\, kg$ block moves towards a light spring with a velocity of $8\, m/s$. When the spring is compressed by $3\, m$,its momentum becomes half of the original momentum. The spring constant of the spring is:

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