Question

0.200 kg stone rests on a frictionless horizontal surface. A buffet of mass 8.60 g traveling horizontally at 360 m/s. takes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s. 

Compute the magnitude of the scene affect it is struck

Compute the direction of the velocity of the stone after it is struck 

Is the collision perfectly?

Answer 1

Part A)

m = mass of bullet = 8.50 g = 0.0085 kg

M = mass of stone = 0.2 kg

v_i = initial velocity of bullet = 360 i^ m/s

V_i = initial velocity of stone = 0

v_f = final velocity of bullet = 250 j^ m/s

V_f = final velocity of stone

using conservation of momentum

m v_i + M V_i = m v_f + M V_f

(0.0085) (360 i^) + (0.2) (0) = (0.0085) (250 j^) + (0.2) V_f

3.06 i^ - 2.125 j^ = (0.2) V_f

V_f = 15.3 i^ - 10.625 j^

magnitude is given as

|V_f| = sqrt((15.3)² + (- 10.625)²) = 18.6 m/s

b)

direction is given as

= tan^-1(V_fy/V_fx) = tan^-1(10.625/15.3) = 34.76 below the initial horizontal direction of bullet

c)

KE_i = initial total KE = (0.5) (m vi² + MVi²) = (0.5) ((0.0085)(360)² + (0.2) (0)²) = 550.8 J

KE_f = final total KE = (0.5) (m v_f² + MV_f²) = (0.5) ((0.0085)(250)² + (0.2) (18.6)²) = 300.22 J

since final and initial Total kinetic energies are not same , hence collision is not perfectly elastic