## College Physics (4th Edition)

(a) The acceleration of the pumpkin is $0.34~m/s^2$ down the slope. The acceleration of the watermelon is $0.34~m/s^2$ up the slope. (b) The pumpkin moves a distance of 1.5 cm (c) The watermelon's speed is 6.8 cm/s
(a) The component of the pumpkin's weight directed down the slope pulls the system toward the left. The component of the watermelon's weight directed down the slope pulls the system toward the right. We can find the acceleration of the system toward the left: $\sum F = (m_p+m_w)~a$ $m_p~g~sin~53^{\circ}-m_w~g~sin~30^{\circ} = (m_p+m_w)~a$ $a = \frac{m_p~g~sin~53^{\circ}-m_w~g~sin~30^{\circ}}{m_p+m_w}$ $a = \frac{(7.00~kg)(9.80~m/s^2)~sin~53^{\circ}-(10.0~kg)(9.80~m/s^2)~sin~30^{\circ}}{(7.00~kg)+(10.0~kg)}$ $a = 0.34~m/s^2$ The acceleration of the pumpkin is $0.34~m/s^2$ down the slope. The acceleration of the watermelon is $0.34~m/s^2$ up the slope. (b) We can find the distance the pumpkin travels in $0.30~s$: $d = \frac{1}{2}at^2$ $d = \frac{1}{2}(0.34~m/s^2)(0.30~s)^2$ $d = 0.015~m = 1.5~cm$ The pumpkin moves a distance of 1.5 cm (c) We can find the watermelon's speed after $0.20~s$: $v_f = v_0+at$ $v_f = 0+(0.34~m/s^2)(0.20~s)$ $v_f = 0.068~m/s = 6.8~cm/s$ The watermelon's speed is 6.8 cm/s