A 20 N force (not shown) acts on the 2-kg block until it strikes the 1-kg ball i

Question

A 20 N force (not shown) acts on the 2-kg block until it strikes the 1-kg ball initially at rest and hanging from a cord attached at 0. When the block strikes the ball, it is moving with a velocity v_0 = 3 m/s. After the impact the ball swings away and block continues to move for some distance x before stop as shown. The coefficient of restitution e between the block and the sphere is 0.6. Determine The height h reached by the ball after the impact. The velocity of the ball immediately after the impact. The distance x travelled by the block after impact. The time the constant force acting on the block before impact

Explanation / Answer

solution:

1)for all collision law conservation of momentum is valid and certain loss of kinetic energy takes place when e<1

2)here for given system ,law of conservation of momentum is

m1*u1+m2*u2=m1*v1+m2*v2

u2=0

where C.R.=v2-v1/u1

v2-v1=1.8

v2=v1+1.8

on putting we get that

we get that

v1=1.4 m/s

and v2=3.2 m/s

2)as for this loss of kinetic energy occure

K.E.lost=K.E.u-K.E.v=[.5*m1*u^2]-.5[m1*v1^2+m2*v2^2]=1.92 j

this K.E. lost is due to friction and hence frictional work done is

Wd=F*d

F=mu*m1*g=.75*2*9.81=14.715 N

Wd=K.E. lost=1.92=14.715*d

d=.1304 m

hence distance move before collision is

d=.1304 m

V=3

u=0

hence accelaration by motion equation

u1^2=uo^2+2*a*d

a=34.4882 m/s2

hence time when constant force applied to accelarate system is

u1=Uo+at

time=u1/a=.08698 sec

4)height reached by ball is

h=v2^2/2*g=.5219 m

5)distance move by block is

K.E. of block=friction loss

.5*m1*v1^2=14.715*d1

d1=.1331 m

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