Question

16 As shown in Figure (5), block A, with a mass of 10 kg, slides down the slope from stationary to hit block B, with a mass of 5 kg. Block A travels 5 m immediately before hitting block B. The sliding friction coefficient between the blocks and the slope is = 0.3. O-x-y is a sta- tionary inertial reference frame. The slope angle, 0-45°. What is the speed of block A immediately before hitting block B with respect to 0-x-y? (A) between 5.0 and 5.5 m/s. (B) between 6.0 and 6.5 m/s. (C) larger than 6.5 m/s. 17 As shown in Figure (5), block A, with a mass of 10 kg, slides down the slope from stationary to hit block B, with a mass of 5 kg. The sliding friction coefficient between the blocks and the slope is Ha=0.3. 0-x-y is a stationary inertial reference frame. The slope angle, 0 = 45°. If the speed of block A was measured to be 5.4 m/s just before hitting block B. What is the speed of block B immediately before impact observed by the mass center of block A and block B, defined during the time period between immediately before and after the impact? (A) between 1.0 to 1.75 m/s, moving up the slope. (B) between 3.5 to 4.0 m/s, moving up the slope. (C) larger than 4.0 m/s, moving up the slope. 18 Per Problem #17, how much total energy is lost due to the impact if block A is found to be traveling down the slope at 2.7 m/s with respect to 0-x-y immediately after the impact? The total energy loss refers to the total energy loss for the system of blocks A and B. (A) between 30 to 40 Joules. (B) between 60 to 80 Joules. (C) larger than 100 Joules. 19 Per Problem #17, which of the following statements regarding the energy loss observed by the mass center reference frame and O-x-y is true? (A) The energy loss observed by O-x-y is larger than that by the mass center frame (B) The energy loss observed by O-x-y is lower than that by the mass center frame (C)The energy loss observed by 0-x-y is the same as that by the mass center frame

Answer 1

Just apply D Alemberts principle for block A that is dynamic equilibrium.

IN ma Solution MA= lokg , MB = 5kg. dist. travelled dA=5m before hitting ble B M=0:3 w 450 0=45° Consider dyanamic egum for big. A block A. Note'- consider along Ety=0 incline line as a & 1 N = log sinas lor to it as y dirn, i {fx-man=0 - PR + 'log los 45-lodge =0 109x = 10x9.81 cos 45-013X10X9.815645 I ax=4.8557 m/ge we have s=utt tat² @ v²=4²+zas (I v-u=at (1) I 5= € 4.8551xt ( U=O oes A Starts from lit = 1.435 See stationary N-0=418557811435 V = 6.968 m/s - Speed of Block A before hitting BL-B = 6.968m/s