Question

A wooden block is cut into two pieces; one is 3 times heavier than the other. A depression is made in both pieces; a firecracker is inserted in the depression and the block is reassembled. The reassembled block is placed on a rough surface and the fuse on the firecracker is lit. When the firecracker explodes, the two pieces of the block slide apart. What is the ratio of the distances traveled by each block?

Answer 1

Answer:

Let A represent the more massive piece, and B the less massive piece.  Thus

Ma = 3 Mb

In the explosion, momentum is conserved.  We have

Va = V2 = 0

Pin = Pfinal

0 = MaVa + Mb Vb

= 3MbVa + Mb Vb

Va = - 1/3 x Vb

For each block, the kinetic energy gained during the explosion is lost to negative work done by friction on the block.

Wfr = KEf - KEi = -1/2 mbv^2

But work is also calculated in terms of the force doing the work and the distance traveled.

Mfr = Ffr delta x cos 180 degrees

= -Uk Fn x delta x = -Ukmg x delta x

Equate the two work expressions, solve for the distance traveled, and find the ratio of distances.

_-1/2mv^2 = -Uk mg x delta x

Delta x = v^2 / gUx

Delta Xa / delta Xb = (- 1/3 x Vb)^2 / Vb ^2 = 1/9

And so

(Delta x)heavy / (delta x ) light = 1/9

Answer 2

The energy and momentum of the sliding blocks are assumed to be conserved

• The ratio of the distances traveled by the blocks m₁, and m₂, is 1 : 9

Reason:

Known parameters;

Mass of one piece of the wooden block, m₁ = 3 × Mass of other piece, m₂

Location of firecracker = Depression made in the pieces

Required:

The ratio of the distances traveled by each block when the fire cracker slide apart

Solution:

Let v₁ and v₂ represent the speed with which the blocks m₁ and m₂ slide apart, we have;

(m₁ + m₂) × v = m₁·v₁  + m₂·v₂

The initial velocity of the blocks, v = 0, therefore;

(m₁ + m₂) × 0 = 0 = m₁·v₁  + m₂·v₂

m₁·v₁  = -m₂·v₂

m₁ = 3·m₂, therefore;

3·m₂ × v₁ = -m₂·v₂

3·v₁ = -v₂

Distance = Velocity, v × Time, t

The distance traveled by each block in the same time, t, d₁, and d₂, are given as follows;

The kinetic energy of the block, m₁, K.E.₁ = 0.5·m₁·v₁²

Work done by friction for block, m₁ = μ·m₁·g·d₁

The kinetic energy of the block, m₂, K.E.₂ = 0.5·m₂·v₂²

Work done by friction for block, m₂ = μ·m₂·g·d₂

Where;

μ = Coefficient of dynamic friction

Therefore, by the law of conservation of energy, we have;

0.5·m₁·v₁² = μ·m₁·g·d₁

0.5·m₂·v₂² = 0.5·m₁/3·(-3·v₁)² = 3·0.5·m₁·v₁²

$\dfrac{1}{2} \cdot m_2 \cdot v_2^2 = \dfrac{1}{2} \cdot \dfrac{m_1}{3} \cdot 9 \cdot v_1^2 = \dfrac{3}{2} \cdot {m_1} \cdot v_1^2$

$\dfrac{3}{2} \cdot {m_1} \cdot v_1^2 = \mu \cdot m_2 \cdot g \cdot d_2 = \dfrac{\mu}{3} \cdot m_1 \cdot g \cdot d_2$

From 0.5·m₁·v₁² = μ·m₁·g·d₁, we have;

$\dfrac{3}{2} \cdot {m_1} \cdot v_1^2 = 3 \cdot \mu \cdot m_1 \cdot g \cdot d_1$

Therefore;

$3 \cdot \mu \cdot m_1 \cdot g \cdot d_1 = \dfrac{\mu}{3} \cdot m_1 \cdot g \cdot d_2$

9·d₁ = d₂

$\dfrac{d_2}{d_1} = \dfrac{9}{1}$

Therefore, the ratio of the distances the blocks m₁, and m₂ travels, which is d₁ : d₂ = 1 : 9

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