A LEVEL PROBLEMS

1)The breaking strength of bone is 170 x 10⁶ N/m² (force per unit area), and its smallest cross-sectional area is 2.5 x 10^-4 m². When a person jumps from a certain height, the person usually breaks his or her fall by bending a certain distance. From what maximum height can a 75-kg person jump without breaking the lower leg bone on either leg? Ignore air resistance and assume that the center of mass moves a distance of 0.60-m in breaking the fall.

The maximum force that each leg can exert without breaking is the breaking strength times the cross-sectional area of the bone.

        (170 ´ 10⁶ N/m²)(2.5 ´ 10^–4 m²) = 4.25 ´ 10⁴ N,

So, if there is an even landing with both feet, the maximum force allowed on the body is 8.50 ´ 10⁴ N.

We use the work-energy principle for the fall to find the landing speed:

        0 = Dk + DU; 

        0 = ½mv_land² – 0 + (0 – mgh_max), or v_land² = 2gh_max.

The impulse from the maximum force changes the momentum on landing. If we take down as the positive direction and assume the landing lasts for a time t, we have

        – F_maxt = mDv = m(0 – v_land),or t = mv_land/F_max.

We have assumed a constant force, so the acceleration will be constant. For the landing we have

        y = v_landt + ½at² = v_land(mv_land/F_max) + ½ (– F_max/m)(mv_land/F_max)² = ½mv_land²/F_max = mgh_max /F_max ;

        0.60 m = (75 kg)(9.80 m/s²)h_max/(8.50 ´ 10⁴ N), which gives h_max = 69 m.

2)In a physics lab, a small cube slides down a frictionless incline as shown in the figure, and elastically collides with a cube at the bottom (the cube at the bottom has half of the mass of the sliding cube). If the incline is 30-cm high and the table is 90 cm off the floor, where does each cube land?

We find the speed after falling a height h from energy conservation:

        ½Mv² = Mgh,or v = (2gh)^1/2.

The speed of the first cube after sliding down the incline and just before the collision is

        v₁ = [2(9.80 m/s²)(0.20 m)]^1/2= 1.98 m/s.

For the elastic collision of the two cubes, we use momentum conservation:

        Mv₁ + mv₂ = Mv₁¢ + mv₂¢;

        M(1.98 m/s) + ½M(0) = Mv₁¢ + !Mv₂¢.

Because the collision is elastic, the relative speed does not change:

        v₁ – v₂ = – (v₁¢ – v₂¢),or1.98 m/s – 0 = v₂¢ – v₁¢.

Combining these two equations, we get 

        v₁¢ = 0.660 m/s,andv₂¢ = 2.64 m/s.

Because both cubes leave the table with a horizontal velocity, they will fall to the floor in the same time, which we find from

        H = ½gt²; 0.90 m = ½ (9.80 m/s²)t², which gives t = 0.429 s.

Because the horizontal motion has constant velocity, we have

        x₁ = v₁¢t = (0.660 m/s)(0.429 s) =0.28 m;

        x₂ = v₂¢t = (2.64 m/s)(0.429 s) =1.1 m.

3)A 5.0-kg body moving in the x-direction at 5.5 m/s collides head-on with a 3.0-kg body moving in the –x-direction at 4.0 m/s. Find the final velocity of each mass if: (a) the masses stick together; (b) if the collision was elastic; (c) if the 5.0 kg body is at rest after the collision; (d) if the 3.0-kg body is at rest after the collision; (e) if the 5.0-kg body has a velocity of 4.0 m/s in the –x-direction after the collision. Are the results in (c), (d), and (e) “reasonable”?

For the momentum conservation of this one-dimensional collision, we have

m₁v₁ + m₂v₂ = m₁v₁¢ + m₂v₂¢.

 
(a)If the bodies stick together, v₁¢ = v₂¢ = V:

        (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg + 3.0 kg)V, which gives V =v₁¢ = v₂¢ = 1.9 m/s.

 
(b)If the collision is elastic, the relative speed does not change:

        v₁ – v₂ = – (v₁¢ – v₂¢),or5.5 m/s – (– 4.0 m/s) = 9.5 m/s = v₂¢ – v₁¢.

The momentum equation is

        (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg)v₁¢ + (3.0 kg)v₂¢,or

        (5.0 kg)v₁¢ + (3.0 kg)v₂¢ = 15.5 kg·m/s.

When we combine these two equations, we get v₁¢ = – 1.6 m/s, v₂¢ = 7.9 m/s.

 
(c)If m₁ comes to rest, v₁¢ = 0.

        (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = 0 + (3.0 kg)v₂¢, which gives v₁¢ = 0, v₂¢ = 5.2 m/s.

 
(d)If m₂ comes to rest, v₂¢ = 0.

        (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg)v₁¢ + 0, which gives v₁¢ = 3.1 m/s, v₂¢ = 0.

 
(e)The momentum equation is

        (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg)(– 4.0 m/s) + (3.0 kg)v₂¢, 

which gives v₁¢ = – 4.0 m/s, v₂¢ = 12 m/s.

The result for (c) is reasonable. The 3.0-kg body rebounds.

The result for (d) is not reasonable. The 5.0-kg body would have to pass through the 3.0-kg body.

To check the result for (e) we find the change in kinetic energy:

        Dke= (½m₁v₁¢² + ½m₂v₂¢²) – (½m₁v₁² + ½m₂v₂²)

        = ½ [(5.0 kg)(5.5 m/s)² + (3.0 kg)(– 4.0 m/s)²] – ½ [(5.0 kg)(– 4.0 m/s)² + (3.0 kg)(12 m/s)²]

        = + 156 J.

Because the kinetic energy cannot increase in a simple collision, the result for (e) is not reasonable.

4)An explosion breaks an object, originally at rest, into two fragments. One fragment has twice the kinetic energy of the other. What is the ratio of their masses?

Momentum conservation gives 

        0 = m₁v₁¢ + m₂v₂¢, or v₂¢/v₁¢ = – m₁/m₂.

The ratio of kinetic energies is

        ke₂/ke₁ = ½m₂v₂¢² /½m₁v₁¢² = (m₂/m₁)(v₂¢/v₁¢)² = 2.

When we use the result from momentum, we get

        (m₂/m₁)(– m₁/m₂)² = 2, which gives m₁/m₂ = 2.