CP? In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s (?Fig. P8.87?). Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 37o from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

Solution 95P This question can be solved considering the equations of conservation of energy and momentum. Given, mass of the package m = 15p0 kg Speed of the package v = 3p00 m/s Height of the chute above the bottom of the cart h = 4.00 m Let v b1 the speed of the package when it lands in the cart. Therefore, using the principle of conservation of energy, (a) Total initial energy = Total final energy 1 2 1 2 2m vp p + m gp = m v2 p 1 2 2 v1 = v p + 2gh v1 2 = (3.00) + 2 × 9.8 × 4 m /s2 2 2 2 2 v1 = 9.00 + 78.4 m /s v1= 9.35 m/s The speed of the package when it lands in the cart is 9.35 m/s. (b) The cart is moving to the left with speed of 5.00 m/s. So, v =c 5.00 m/s Mass of the cart m = c0.0 kg 0 The package slides down the inclined plane. The angle of inclination is 37 C . 0 Therefore, its x-component of speed is v px= 3.00 m/s × cos 37 = 2.39 m/s Initial momentum of the system, P = 15.0 kg × 2.39 m/s + 50.0 kg × ( 5.00 m/s) = 214.15 kg.m/s i If the final speed of the cart is v2. Final momentum of the system P = (15 + 50)v = 65v f 2 2 Therefore, P = Pi f 214.15 kg.m/s = 65v 2 v =2 3.29 m/s The speed of the cart would be 3.29 m/s. The negative sign indicates that the cart would move to the left.