Answer
(a) Block R travels a distance of $~~1.92~m$
(b) Block R travels a distance of $~~0.64~m$
Work Step by Step
(a) By conservation of momentum, the momentum of each block must be equal in magnitude after being released.
We can find the speed of block $R$ after being released:
$p_R = p_L$
$m_R~v_R = m_L~v_L$
$v_R = \frac{m_L~v_L}{m_R}$
$v_R = \frac{(1.00~kg)~(1.20~m/s)}{(0.500~kg)}$
$v_R = 2.40~m/s$
We can find the distance that block $R$ travels in $0.800~s$:
$d = (2.40~m/s)(0.800~s) = 1.92~m$
Block R travels a distance of $~~1.92~m$
(b) Let the velocity of block $R$ be $v_R$
Let the velocity of block $L$ be $~~v_R-1.20~m/s$
We can use conservation of momentum to find the velocity of block $R$ after being released:
$p_f = p_i$
$m_L~v_L + m_R~v_R = 0$
$m_L~(v_R-1.20~m/s) + m_R~v_R = 0$
$(m_L+m_R)~(v_R) = (m_L)(1.20~m/s)$
$v_R = \frac{(m_L)(1.20~m/s)}{m_L+m_R}$
$v_R = \frac{(1.00~kg)(1.20~m/s)}{1.00~kg+0.500~kg}$
$v_R = 0.800~m/s$
We can find the distance that block $R$ travels in $0.800~s$:
$d = (0.800~m/s)(0.800~s) = 0.64~m$
Block R travels a distance of $~~0.64~m$
