Impulse, Momentum, Collisions Problems and Solutions

 Problem#1

A bullet of mass m is fired into a block of mass M initially at rest at the edge of a frictionless table of height h (Fig. 1). The bullet remains in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

Fig.1

Answer:
Using conservation of momentum from just before to just after the impact of the bullet with the block:

mv_i = (M + m)v_f

v_i = v_f (M + m)/m                                          (1)

The speed of the block and embedded bullet just after impact may be found using kinematic equations:

d = v_ft and h = ½ gt²

Thus, v_f = d[2h/g]^1/2                       (2)

Substituting (2) into (1) from above gives

v_f = (M + m)d[g/2h]^1/2/m

Problem#2
A 0.500-kg sphere moving with a velocity (2.00i - 3.00j + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with a velocity (-1.00i + 2.00j - 3.00k) m/s. (a) If the velocity of the 0.500-kg sphere after the collision is (-1.00i + 3.00j + 8.00k) m/s, find the final velocity of the 1.50-kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic). (b) If the velocity of the 0.500-kg sphere after the collision is (-0.250i + 0.750j - 2.00k) m/s, find the final velocity of the 1.50-kg sphere and identify the kind of collision. (c) What If? If the velocity of the 0.500-kg sphere after the collision is (-1.00i + 3.00j + ak) m/s, find the value of a and the velocity of the 1.50-kg sphere after an elastic collision.

Answer:
(a) Conservation of momentum given by

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f or

(0.500kg)(2.00i - 3.00j + 1.00k) m/s + (1.500kg)(-1.00i + 2.00j - 3.00k) m/s = (0.500kg)(-1.00i + 3.00j + 8.00k) m/s + (1.500kg)v_2f

Thus, v_2f = 0

The original kinetic energy is

K_i = ½m₁v_1i² + ½m₂v_2i²

K_i = ½ (0.500kg)(2² + 3² + 1²)m²/s² + ½ (1.500kg)(1² + 2² + 3²)m²/s²

K_i = 14.0 J

and the final kinetic energy is

K_f = ½m₁v_1f² + ½m₂v_2f²

K_f = ½ (0.500kg)(1² + 3² + 8²)m²/s² + 0 = 18.5J

different from the original energy so the collision is inelastic

(b) We follow the same steps as in part (a)

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f or

(0.500kg)(2.00i - 3.00j + 1.00k) m/s + (1.500kg)(-1.00i + 2.00j - 3.00k) m/s = (0.500kg)(-0.250i + 0.750j - 2.00k) m/s + (1.500kg)v_2f

v_2f = (-0.250i + 0.750j - 2.00k) m/s

We see v_2f = v_1f, so the collision is perfectly inelastic.

(c) Conservation of momentum given by

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f or

(0.500kg)(2.00i - 3.00j + 1.00k) m/s + (1.500kg)(-1.00i + 2.00j - 3.00k) m/s = (0.500kg)(-1.00i + 3.00j + ak) m/s + (1.500kg)v_2f

v_2f = (-2.67 – 0.333a)k m/s

Conservation of momentum given by

½m₁v_1i² + ½m₂v_2i = ½m₁v_1f² + ½m₂v_2f²

14.0J = ½ (0.500kg)(1² + 3² + a²)m²/s² + ½ (1.50kg)(2.67 + 0.333a)²m²/s²

14.0J = 2.5J + 0.250a² + 5.33J + 1.33a + 0.0833a²

0.333a² + 1.33a – 6.17 = 0

Then,
a = {–1.33 ± [1.33² – 4(0.333)(–6.17)]^1/2}/0.667

So, a = 2.74, v_2f = (-2.67 – 0.333(2.74))k = –3.58k m/s

So, a = –6.74, v_2f = (-2.67 – 0.333(–6.74))k = –0.419k m/s

Problem#3
A small block of mass m₁ = 0.500 kg is released from rest at the top of a curve-shaped frictionless wedge of mass m₂ = 3.00 kg, which sits on a frictionless horizontal surface as in Figure 2a. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure 2b. (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge?

Answer:
(a) The initial momentum of the system is zero, which remains constant throughout the motion.
Therefore, when m₁ leaves the wedge, we must have

m₁v_block + m_2­v_wedge = 0 or

(0.500kg)(+4.00m/s) + (3.00kg)v_wedge = 0

So, v_wedge = –0.667 m/s

(b) Using conservation of energy for the block-wedge Earth system as the block slides down the smooth (frictionless) wedge, we have

(K_block + U_system)_i + (K_wedge)_I = (K_block + U_system)_f + (K_wedge)_f

0 + (0.500kg)(9.80 m/s²)h + 0 = ½ (0.500kg)(4.00m/s)² + 0 + ½ (3.00kg)(–0.667 m/s)²

h = 0.952 m

Problem#4
A bucket of mass m and volume V is attached to a light cart, completely covering its top surface. The cart is given a quick push along a straight, horizontal, smooth road. It is raining, so as the cart cruises along without friction, the bucket gradually fills with water. By the time the bucket is full, its speed is v. (a) What was the initial speed v_i of the cart? Let ρ represent the density of water. (b) What If? Assume that when the bucket is half full, it develops a slow leak at the bottom, so that the level of the water remains constant thereafter. Describe qualitatively what happens to the speed of the cart after the leak develops.

Answer:
(a) Conservation of the x component of momentum for the cart-bucket-water system:
mv_i + 0 = (m + ρV)v

v_i = (m + ρV)v/m

(b) Raindrops with zero x-component of momentum stop in the bucket and slow its horizontal
motion. When they drip out, they carry with them horizontal momentum. Thus the cart
slows with constant acceleration.  

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