An incident ball A of mass .20 kg is sliding at 2.3 m/s on horizontal tabletop I

Question

An incident ball A of mass .20 kg is sliding at 2.3 m/s on horizontal tabletop It makes head-on collision with a target ball B of mass 0.60 kg at rest at edge of table. As result of collision, the incident ball rebounds, sliding backwards at 1.0 m/s immediately after the collision

Calculate spped of the .60 kg target ball immediately after collision

The tabletop is 1.20 m above level, horizontal floor. Target ball is projected horizontally and and initially strikes floor at a horizontal displacement d from the point of collision Calculate horizontal displacement d of the target ball

Explanation / Answer

Assumuning Elastic collision,

for conservation of momentum

m1u1+m2u2= m1v1+m2v2. m1 = 0.20 kg, m2=0.60kg u1=2.3m/s, u2=0 m/s; v1=-1m/s

(0.20*2.3) + 0 = (0.20*-1)+0.6*(v2)

0.6v2 = 0.66

v2 = 1.1 m/s

Since the ball is horizontally projected. In vertical direction, the initial velocity = 0;

hence y = 1/2 at2. y = 1.2m, a = g = 9.8;

t = sqrt(2*y/a) = 0.495 sec

In horizontal direction.

x = ut since a=0;

= 1.1 x 0.495

= 0.5445 m = d

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