The 2-kg ball A is thrown so that it travels horizontally at 10 m/s when it strikes the 6-kg block B as it is traveling down the incline at 1 m/s. If the coefficient of restitution between the ball and the block is e=0.6, determine the speeds of the ball and the block just after impact.

Answer

The velocities of A and B after impact are -0.8m/s and 4.6m/s respectively

Explanation:

Let Va1 speed of A before collission=10m/s

Va2=speed after collission=?

Also Vb1=1m/s

Vb2=?

Mass of A is Ma=2kg.

Mass of B is Mb=6kg

coefficient of restitution e=0.6

From linear momentum conservation

Ma*Va1+Mb*Vb1=Ma*Va2+Mb*Vb2

2*10+6*1=2Va2+6Vb2

26=2(Va2+3vb2)

Va2=13-3vb21.

Also from e=(vb2-va2)/(va1-vb1)

0.6=vb2-va2/10-1

5.4=vb2-va2

So Va2=vb2-5.4---2

Substitute equation 2 into 1

13-3vb2=vb2-5.4

Solving for Vb2= 4.6m/s

And Va2=vb2-5.4= 4.6-5.4=-0

8m/s

v=ir = 3x15 =45 v

d. air pressure decreases as altitude increases.