Question

PART A Hardcopy due 11 am Moody December 12 A m, 500 object released with an iniull speed o120 oms from the top of a Iough track im above p a table which is h-2m high is desetmised that friction does 5l af wark alung the trakk lo the bullum. It these collides ouple with a m 1kg block that is initially at rest on Lable as shown. (a) Find the veiociy of the coupled object just after the collision (bU What percentage of the kinetic energy is "lost" as result of the oollision? (c) how away from the bersam of the tahle does the coupled mass land? (d) with what speed (ie the magunadeed the velocity vector and angle hit the tloor?

how do I factor in collision and friction

Answer 1

Consider the motion on the rough track

W_f = work done by friction = 5 J

V_i = initial speed at the top = 20 cm /s = 0.2 m/s
V_f = final speed at the bottom = ?

h = height of track = 3 m

using conservation of energy between Top and bottom track

(0.5) m V_i² + mg h = (0.5) m V_f² + W_f

(0.5) (0.5) (0.2)² + (0.5) (9.8) (3) = (0.5) (0.5) V_f² + 5

V_f = 6.23 m/s

V = velocity of combination after collision

M = mass of block = 1 kg

using conservation of momentum

m V_f = (m + M) V

(0.5) (6.23) = (0.5 + 1) V

V = 2.08 m/s

b)

KE lost = (0.5) m V_f² - (0.5) (m + M) V² = (0.5) (0.5) (6.23)² - (0.5) (0.5 + 1) (2.08)² = 6.5 J

c)

consider the vertical motion of the coupled mass

V_oy = initial velocity = 0 m/s

a = acceleration = - 9.8

t = time

Y = vertical displacement = - 2m

using the equation

Y = V_oy t + (0.5) a t²

- 2 = 0 t + (0.5) (-9.8) t²

t = 0.64 sec

along the horizontal direction

V_ox = initial velocity along horizontal direction =

X = V_ox t = 2.08 x 0.64 = 1.33 m