1)A child on a boat throws a 5.40-kg package out horizontally with a speed of 10.0 m/s. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg and the mass of the boat is 55.0 kg.

The initial momentum of the system is zero; the boat, child, and package are all at rest.

Once the package is thrown, the boat and child move with a single velocity while the package moves with a velocity of qo m/s.

        0 = (m_boat + m_child)v_boat+ m_packagev_package;

        0 = (55.0 kg + 20.0 kg)v_boat + (5.40 kg)(10.0 m/s), which gives

        v_boat=– 0.720 m/s (opposite to the direction of the package).

2)A 9500-kg boxcar travels on a level frictionless track with a constant speed of 18.0 m/s. A 5750-kg additional load is dropped onto the car. What is the new speed of the car after this additional load was dropped?

For the one-dimensional motion, we take the direction of the first car for the positive direction.

For this perfectly inelastic collision, we use momentum conservation:

        M₁v₁ + M₂v₂ = (M₁ + M₂)V;

        (9500 kg)(18 m/s) + 0 = (9500 kg + 5750 kg) V, which gives V = 11.2 m/s

3)A 15-g bullet strikes and becomes embedded in a 1.10-kg block of wood on a horizontal surface right in front of the gun. If the coefficient of kinetic friction between the wood and the surface is equal to 0.25, and the impact drives the block a distance of 9.5 m before it comes to rest, what was the muzzle speed (initial speed) of the bullet?

On the horizontal surface after the collision, the normal force is F_N = (m + M)g. This problem can be solved by looking at the work done by friction in slowing the object to a stop, and by using conservation of momentum during the collision. We find the common speed of the block and bullet immediately after the embedding by using the work-energy principle for the sliding motion:

        W_fr = Dk;

        – µ_k(m + M)gd = 0 – ½ (M + m)V²;

        0.25(9.80 m/s²)(9.5 m) = ½V², which gives V = 6.82 m/s.

For the collision, we use momentum conservation:

        mv + 0 = (M + m)V;

        (0.015 kg)v = (0.015 kg + 1.10 kg)(6.82 m/s), which gives v = 5.1 ´ 10² m/s.

4)An atomic nucleus initially at rest radioactively decays into an alpha particle and a smaller nucleus. What will be the speed of the recoiling nucleus if the speed of the alpha particle is 3.8 x 10⁵ m/s? Assume that the recoiling nucleus has a mass 57 times greater than that of the alpha particle.

The new nucleus and the alpha particle will recoil in opposite directions.

Momentum conservation gives us

        0 = MV – m_av_a,

        0 = (57m_a)V – m_a(3.8 ´ 10⁵ m/s),which gives V= 6.7 ´ 10³ m/s.

5)A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 5.0 x 10^-3 s. Find (a) the impulse imparted to the golf ball and (b) the average force exerted on the ball by the golf club.

(a)We find the impulse on the ball from

        Impulse = Dp = m Dv =(0.045 kg)(45 m/s – 0) = 2.0 N·s.

        F = Impulse/Dt = (2.0 N·s)/(5.0 ´ 10^–3 s) = 4.0 ´ 10² N.

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6)A ball of mass 0.440 kg moving east with a speed of 3.70 m/s collides head-on with a 0.220-kg mass at rest. If the collision is perfectly elastic, what is the speed and direction of each ball after the collision?

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

        m1v1 + m2v2 = m1v1¢ + m2v2¢;

        (0.440 kg)(3.70 m/s) + (0.220 kg)(0) = (0.440 kg)v1¢ + (0.220 kg)v2¢.

        1.628 kg m/s = (0.440 kg)v1¢ + (0.220 kg)v2¢

Because the collision is elastic, kinetic energy is constant:

        ½ m₁ v²₁ + ½m₂ v²₂ = ½ m₁v₁’² + ½ m₂v₂^’2,

        ½ (0.440 kg)(3.70 m/s)² + 0 = ½ (0.440 kg)v₁’² + ½ (0.220 kg)v₂^’2

        6.02 kg m²/s² = (0.440 kg)v₁’² + (0.220 kg)v₂^’2

Combining these two equations, we get

        v1¢ = 1.23 m/s,andv2¢ = 4.93 m/s.

7)An explosion breaks an object into two pieces one of which has 1.5 times the mass of the other. If 7500 J were released during the explosion, how much kinetic energy did each piece acquire?

Momentum conservation gives 

        0 = m₁v₁¢ + m₂v₂¢;

But m₂ = 1.5 m₁, therefore:

        0 = m₁v₁¢ + 1.5m₁v₂¢,or

        v₁¢ = – 1.5v₂¢.

The kinetic energy of each piece is

        k₂ = ½m₂v₂¢²;

        k₁ = ½m₁v₁¢² = ½ (m₂/1.5)(– 1.5v₂¢)² = (1.5) ½m₂v₂¢² = 1.5k₂ .

The energy supplied by the explosion produces the kinetic energy:

        e = k₁ + k₂ = 2.5k₂;

        7500 J = 2.5k₂ , which gives k₂ = 3000 J.

For the other piece we have

        k₁ = e – k₂ = 7500 J – 3000 J = 4500 J.

Thus 

        k(heavier) = 3000 J; k(lighter) = 4500 J.

8)A 1.0 x 10³ kg Toyota collides into the rear end of a 2.2 x 10² kg Cadillac stopped at a red light. The bumpers lock, the breaks are locked, and the two cars skid forward 2.8 m before stopping. The police officer, who is aware that the coefficient of friction between the tires and road is 0.40, calculates the speed of the Toyota at impact. What was the speed?

On the horizontal surface after the collision, the normal force on the joined cars is F_N = (m + M)g.

We find the common speed of the joined cars immediately after the collision by using the work-energy principle for the sliding motion:

        W_fr = DK;

        – µ_k(m + M)gd = 0 – ½ (M + m)V²;

        0.45(9.80 m/s²)(2.8 m) = ½V², which gives V = 4.68 m/s.

For the collision, we use momentum conservation:

        mv + 0 = (m + M)V;

        (1.0 ´ 10³ kg)v = (1.0 ´ 10³ kg + 2.2 ´ 10³ kg)(4.68 m/s), which gives v = 15 m/s (54 km/h).

9)Four masses m₁ =2.0 kg, m₂ = 1.5 kg, m₃ = 1.0 kg, m₄ = 0.5 kg, are at the positions r₁ = (0.0,0.0) m, r₂ = (2.0, 0.0) m, r₃ = (-2.0,-1.0) m, r₄= (1.0,-1.0) m respectively, on the xy-plane. Find the coordinates of the center of mass of this system.

        X(cm) = Sm_ix_i/Sm_i = [(2.0 kg)(0.0) + (1.5 kg)(2.0 m) + (1.0 kg)(-2.0 m) + (0.5 kg)(1.0 m)]/(2.0 + 1.5 + 1.0 + 0.5) kg

        = [0 + 3.0 kg-m – 2.0 kg-m + 0.5 kg-m]/(5.0 kg) = (1.5 kg-m)/(5.0 kg) =0.30 m

        Y(cm) = Sm_iy_i/Sm_i = [(2.0 kg)(0.0) + (1.5 kg)(0.0 m) + (1.0 kg)(-1.0 m) + (0.5 kg)(-1.0 m)]/(2.0 + 1.5 + 1.0 + 0.5) kg

        = [0 + 0 – 1.0 kg-m - 0.5 kg-m]/(5.0 kg) = (-1.5 kg-m)/(5.0 kg) = -0.30 m

        (0.30, -0.30) m

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