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

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft