Solved Numerical Problem No.5 ( Unit-3: Force and Motion); HSSC-1 (Class-11); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

One ball of mass 0.600 kg traveling 9.00 m/s to the right collides head-on elastically with a second ball of mass 0.300 kg traveling 8.00 m/s to the left. After the collision, what are their velocities after collision? (Ans. -2.33 m/s (2.33 m/s to left) and 14.67 m/s (14.76 m/s to right))

Given:

Mass of first ball = m₁ = 0.6 kg
The initial velocity of the first ball to the right u₁ = 9.00 m/s

Mass of the second ball = m₂ = 0.3 kg

Initial velocity of second ball = u₂ = - 8.00 m/s (-ve sign for opposite in direction to u₁)

To Find:

(a) Final velocity of mass m1 after collision = v₁ = ?

(b) Final velocity of mass m2 after collision = v₂ = ?

Solution:

(a) Final velocity of mass m1 after collision = v₁ = ?

The

formula to find the Final velocity of mass m1 is:

v₁ = $\frac{(m ₁ - m ₂) u 1}{m ₁ + m ₂}$ $\frac {(m₁-m₂)u1}{m₁+m₂}$ + $\frac{2 m ₂ u ₂}{m ₁ + m ₂}$ $\frac {2m₂u₂}{m₁+m₂}$

by putting values

v₁ = $\frac{(0.6 k g - 0.3 k g) 9.00 \frac{m}{s}}{0.6 k g + 0.3 k g}$ $\frac {(0.6 kg - 0.3 kg) 9.00 m/s}{0.6 kg + 0.3 kg}$ + $\frac{2 x 0.3 k g x (- 8.00 \frac{m}{s})}{0.6 k g + 0.3 k g}$ $\frac {2x0.3 kg x (-8.00 m/s) }{0.6 kg + 0.3 kg}$

v₁ = $\frac{(0, 6 k g - 0, 3 k g) 9.00 \frac{m}{s}}{0.6 k g + 0.3 k g}$ $\frac {(0,6 kg - 0,3 kg) 9.00 m/s}{0.6 kg + 0.3 kg}$ - $\frac{2 x 0.3 k g x (8.00 \frac{m}{s})}{0.6 k g + 0.3 k g}$ $\frac {2x0.3 kg x (8.00 m/s) }{0.6 kg + 0.3 kg}$

v₁ = 3 m/s - 5.33 m/s

v₁ = - 2.33 m/s -------------Ans.

The -ve sign shows the final velocity of mass m1 is opposite to its initial velocity i.e it is toward the left.

(b) Final velocity of mass m2 after collision = v₂ = ?

The formula to find the Final velocity of mass m₂ is:

v₂ =

\frac{(m ₂ - m ₁) u ₂}{m ₁ + m ₂}

$\frac {(m₂-m₁)u₂}{m₁+m₂}$ +

\frac{2 m ₁ u ₁}{m ₁ + m ₂}

$\frac {2m₁u₁}{m₁+m₂}$

by putting values

v₂ =

\frac{(0.3 k g - 0.6 k g) (- 8.00 \frac{m}{s})}{0.6 k g + 0.3 k g}

$\frac {(0.3 kg - 0.6 kg) (-8.00 m/s)}{0.6 kg + 0.3 kg}$ +

\frac{2 x 0.6 k g x (9.00 \frac{m}{s})}{0.6 k g + 0.3 k g}

$\frac {2x0.6 kg x (9.00 m/s) }{0.6 kg + 0.3 kg}$

v₂ = -

\frac{(0.3 k g - 0.6 k g) 8.00 \frac{m}{s}}{0.6 k g + 0.3 k g}

$\frac {(0.3 kg - 0.6 kg) 8.00 m/s}{0.6 kg + 0.3 kg}$ +

\frac{2 x 0.6 k g x (9.00 \frac{m}{s})}{0.6 k g + 0.3 k g}

$\frac {2x0.6 kg x (9.00 m/s) }{0.6 kg + 0.3 kg}$

v₂ = -(-2.66) m/s + 12.0 m/s

v₂ = 2.66 m/s + 12.0 m/s

v₂ = 14.66 m/s -------------Ans.

The +ve sign shows the final velocity of mass m₂ is opposite to its initial velocity direction i.e it is toward the right.

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