A.) An 87 kg running back moving at 7 m/s makes a perfectly inelastic collision

Question

A.) An 87 kg running back moving at 7 m/s makes a perfectly inelastic collision with a 100 kg linebacker who is initially at rest. What is the speed of the players just after their collision?

____ m/s

B.) A steel ball of mass m1 = 1.1 kg and a cord of length of L = 1.8 m of negligible mass make up a simple pendulum that can pivot without friction about the point O, as in the figure below. This pendulum is released from rest in a horizontal position, and when the ball is at its lowest point it strikes a block of mass m2 = 1.1 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and that the coefficient of kinetic friction between the block and shelf is 0.10.

-How far does the block slide before coming to rest (assuming that the shelf is long enough)?

____ m

Explanation / Answer

A) m1 = 87 Kg,      u1 = 7 m/s

m2 ( mass of linebacker) = 100 kg,              u2( initial velocity of linebacker) = 0 due to at rest.

From conservation of linear momentum,          pi = pf

m1*u1 + m2*u2 = ( m1+m2)*V          ( since collision is perfactly inelastic)

87*7 +100*0 = (87+100)*V

V( speed of players after collision) = 3.25m/s

B) Since for this part fig, is not given. so i'm answering according to that what i'm got from this information

initial velocity of ball by which it is coming towards the block on shelf, By conservation of energy:

P.E ( at heighest point) = K.E ( at lowest point)

m1*g*L = (1/2)*m1*u12

u1 = (2gL)1/2

By formula for final velocity of second object for elastic collision:

V2 ( velocity attain by block after collision) = ( 2*m2 / m1+m2)*u1 + (m2-m1)*u2/ m1+m2

here u2 ( initial velocity of block = 0 due to at rest

V2 = 2*1.1*( 2*9.8*1.8)1/2 / 1.1+1.1

V2 = 5.93 m/s

also   m2*a = -f                     ( a= acceleration of block, f= friction force= uk*R = uk*m2*g)

m2*a = - uk*m2*g

a = - uk*g

By V2 - u2 = 2as                       ( u= initial velocity of block= V2 = 5.93 m/s, V= final velocity=0)

-u2 = - 2*uk*g*s

s ( distance travelled by block) = 5.932 / 2*0.10*9.8

                                              = 18 m

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