Two putty balls of equal mass moving with equal velocity in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a velocity of $45 \sqrt{2} m s^{- 1}$ each, the velocity of their combined mass after collision is

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Solution

The correct option is
B

Initial momentum
$∣ ∣ \to P ∣ ∣ = m 45 \sqrt{2}^i + m 45 \sqrt{2}^j \Rightarrow \to P = m \times 90$

Final momentum $2 m \times V$

By conservation of momentum $2 m \times V = m \times 90$

$∴ V = 45 m / s$

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Two putty balls of equal masses moving with equal velocity in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a velocity of $45 \sqrt{2} m s^{- 1}$ each, the velocity of their combined mass after collision is

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45 \sqrt{2} m s^{- 1}

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Two putty balls of equal mass moving with equal velocity in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a velocity of 45√2 ms−1 each, the velocity of their combined mass after collision is

The correct option is B Initial momentum∣∣→P∣∣=m45√2^i+m45√2^j⇒→P=m×90 Final momentum 2m×V By conservation of momentum 2m×V=m×90 ∴V=45 m/s

Two putty balls of equal mass moving with equal velocity in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a velocity of $45 \sqrt{2} m s^{- 1}$ each, the velocity of their combined mass after collision is

The correct option is
B

Initial momentum
$∣ ∣ \to P ∣ ∣ = m 45 \sqrt{2}^i + m 45 \sqrt{2}^j \Rightarrow \to P = m \times 90$

Final momentum $2 m \times V$

By conservation of momentum $2 m \times V = m \times 90$

$∴ V = 45 m / s$