Chapter 9: Linear Momentum and Collisions Selected Solutions

Selected Solutions

15. The sum of the skaters' momenta must be zero. The set up is similar to Example 9-3. Let m₁ = 45 kg, v₁ = -0.62 m/s and v₂ = 0.89 m/s.

25. (a) Use momentum conservation. Let the subscripts b and B denote the bullet and the block, respectively.

(b) The final kinetic energy is less than the initial kinetic energy is lost to the heating and deformation of the bullet and block.

(c) Initially, only the bullet has kinetic energy. Therefore,

.

Finally, both the bullet and the block have kinetic energy. Therefore,

.

We see that K_f < K_i which verifies the answer to path (b).

29. Let's make the following definitions:

m₁ = the mass of the truck
m₂ = the mass of the car
v₀ = the initial speed of the truck

Use conservation of momentum to find an equation for the final speed of the truck.

There is one equation and two unknowns. Use conservation of kinetic energy to find a second equation relating v_1f and v_2f.

Substitute for v_1f and solve for v_2f.

Substitute for v_2f in the equation for v_1f.

Using the given information, m₁ = 1620 kg, m₂ = 722 kg, and v₀ = 14.5 m/s, the final speeds of the truck and car are: v_truck = 5.56 m/s and v_car = 20.1 m/s.

39. Due to symmetry, X_cm = 0. Therefore, we only need to calculate Y_cm.

Therefore, (X_cm, Y_cm) = (0, 3.6 x 10^-11 m).

55. Use conservation of momentum to determine the horizontal speed of the bullet and block.

m = the mass of the bullet
M = the mass of the block
v = the initial speed of the bullet
v_f = the final speed of the bullet + block

Recall that for a mass that is initially stationary...

We can now substitute for v_f and t to find the horizontal distance:

Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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