Question

12/13/2019 11:00 PM 77/100 12/12/2019 10:48 PM Gradebook Print Calculator Periodic Table Question 31 of 32 и эарту-саттттту macmilan learning Map rout in space, a bored astronaut fires an antique gun at a sack of rocks (mass M). The bullet (mass m) Dounces around inside the sack a few times, eventually emerging with a reduced, but still high, speed. As the bullet flies off into space, the astronaut remembers with dismay that it was a priceless historical artifact. Fortunately, the astronaut knows the initial speed vo of the bullet, and that the sack was initially at rest but after being hit was found to be moving with speed Vs at angle os, as shown in the diagram. In what direction should the astronaut pursue the bullet, assuming that he starts from the initial location of the sack, and what minimum speed vb must he go faster than in order to catch up with it? Use the numerical values given below, and enter the direction as an angle measured clockwise from the x-axis. m= 0.0250 kg M=45.0 kg vo=1550 m/s y = 0.425 m/s 6 = 29.50 before after 1 = ? Number Number Hint Previous Give Up & View Solution Check Answer Next Exit about us careers privacy policy terms of use earch Q & A O w e o

Answer 1

This question is solved by conservation of linear momentum.

Initial total linear momentum in horizontal direction = m v = 0.025 1550 = 38.75 kg m/s ...............(1)

where m is mass of bullet and v is speed of bullet.

Initial total linear momentum in vertical direction = 0 ..................(2)

total linear momentum in vertical direction just after bullet exiting the sack = M v_s sin_s - m v_b sin_b ..........(3)

by conservation of linear momentum along vertical direction, using eqn.(2) and eqn.(3),

we get,     v_b sin_b = ( M / m ) v_s sin_s = ( 45 / 0.025 ) 0.425 sin29.5^o 376.7 m/s .............(4)

total linear momentum in horizontal direction just after bullet exiting the sack = M v_s cos_s + m v_b cos_b ..........(5)

by conservation of linear momentum along horizontal direction, using eqn.(1) and eqn.(5),

M v_s cos_s + m v_b cos_b = 38.75

v_b cos_b = [ 38.75 - ( 45 0.425 cos29.5^o ) ] / 0.025 = 884.18 ..................(6)

By dividing eqn.(4) with eqn.(6) , tan _b = 0.426 , _b = 23^o

Hence, from eqn.(4) , v_b = 376.7 / sin _b = 964 m/s