Problem 11.27 A 69.0 kg football player is gliding across very smooth ice at 2.0

Question

Problem 11.27 A 69.0 kg football player is gliding across very smooth ice at 2.05 m/s . He throws a 0.470 kg football straight forward.

Part A

What is the player's speed afterward if the ball is thrown at 18.0 m/srelative to the ground?

Part B

What is the player's speed afterward if the ball is thrown at 18.0 m/srelative to the player?

(2).The air-track carts in the figure(Figure 1) are sliding to the right at 1.0 m/s. The spring between them has a spring constant of 140 N/m and is compressed 4.4 cm. The carts slide past a flame that burns through the string holding them together.

Part 2

What is the speed of 100-g cart?

What is the speed of 300-g cart?

Explanation / Answer

Given

   Mass of football player M = 69 kg,

   ball mas m = 0.47 kg,
initially both are moving with velocity 2.05 m/s

and after thrown by the player ball moves with velocity 18 m/s

now the momentum conservation

  
initial momentum is P1 = (M+m)v = (69+0.47)2.05 kg m/s = 142.4135 kg m/s

now the ball moving with velocity 18 m/s

   142.4135 = (69*v+0.47*18)

   v = 1.941 m/s

PArtA : Player speed afterward if the ball is thrown at 18 m/s is relative to the ground =1.941 m/s

Part B :

relative velocity of the ball is V = 18+2.05 = 20.05 m/s

now 142.4135 = 69*v+0.470*20.05 ==> v = 1.93 m/s

player's speed afterward if the ball is thrown at 18.0 m/srelative to the player is

   V = 1.93 m/s

2.

air track cart moving to the right at v = 1 m/s

Dx = 4.4 cm = 0.044 m, spring force F = 140 N

initial and final total energy of the system is constnat

that is sum of initial(k.e1+k.e2+ p.e) = final (k.e1+k.e2+p.e)

and assume that the initial velocity of both carts zero and the final dX =0

so the equation is like this

   0.5*m1(v1f)^2 + 0.5m2(v2f)^2 +0 = 0+0+0.5k(dXi)^2

  
from conservation of momentum

   m1v1f +m2v2f = m1v1i +m2v2i
   m1v1f +m2v2 = 0

v1f = -m2(v2f)/m1 ----------(1)

and substituting the value of V1f in the energy equation and solving for v2f

   V2f = sqrt((k(dX)^2)/(m2(1+(m2/m1))))

   v2f = sqrt((140(0.044)^2)/(0.3(1+(0.3/0.1)))) m/s

   v2f = 0.475254322 m/s
and v1f = -0.3(0.47525432)/0.1 m/s

   = -1.42576296 m/s
  
1) 100 g cart will move with speed -1.42576296 m/s
2) 300 g cart will move with speed 0.475254322 m/s

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