Two particles of masses and in projectile motion have velocities and respectively at time . They collide at time . Their velocities become and at time while still moving in air. The value of is [2001]

Question

Two particles of masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}'$ and ${\vec{v}}_{2}'$ at time $2 t_{0}$ while still moving in air. The value of $| (m_{1} {\vec{v}}_{1}' + m_{2} {\vec{v}}_{2}') - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) |$ is [2001]

Smart Achievers · Accepted Answer

(4)

$F_{ext} = (m_{1} + m_{2}) g and Δ t = 2 t_{0}$

$F_{ext} = \frac{Δ p}{Δ t}$

$∴$ $Δ p = F_{ext} Δ t = (m_{1} {\vec{v}}_{1}' + m_{2} {\vec{v}}_{2}') - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})$

$= (m_{1} + m_{2}) g \times 2 t_{0}$

Solution

1 zero

2 $(m_{1} + m_{2}) g t_{0}$

3 $\frac{1}{2} (m_{1} + m_{2}) g t_{0}$

4 $2 (m_{1} + m_{2}) g t_{0}$

Ans.
(4)

$F_{ext} = (m_{1} + m_{2}) g and Δ t = 2 t_{0}$

$F_{ext} = \frac{Δ p}{Δ t}$

$∴$ $Δ p = F_{ext} Δ t = (m_{1} {\vec{v}}_{1}' + m_{2} {\vec{v}}_{2}') - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})$

$= (m_{1} + m_{2}) g \times 2 t_{0}$

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Solution

Q. Two particles of masses m1 and m2 in projectile motion have velocities v→1 and v→2 respectively at time t=0. They collide at time t0. Their velocities become v→1' and v→2' at time 2t0 while still moving in air. The value of |(m1v→1'+m2v→2')-(m1v→1+m2v→2)| is [2001]

1 zero

2 (m1+m2)gt0

3 12(m1+m2)gt0

4 2(m1+m2)gt0

Ans. (4) Fext=(m1+m2)g and Δt=2t0 Fext=ΔpΔt ∴ Δp=Fext Δt =(m1v→1'+m2v→2')-(m1v→1+m2v→2) =(m1+m2)g×2t0

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2 $(m_{1} + m_{2}) g t_{0}$

3 $\frac{1}{2} (m_{1} + m_{2}) g t_{0}$

4 $2 (m_{1} + m_{2}) g t_{0}$

Ans.
(4)

$F_{ext} = (m_{1} + m_{2}) g and Δ t = 2 t_{0}$

$F_{ext} = \frac{Δ p}{Δ t}$

$∴$ $Δ p = F_{ext} Δ t = (m_{1} {\vec{v}}_{1}' + m_{2} {\vec{v}}_{2}') - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})$

$= (m_{1} + m_{2}) g \times 2 t_{0}$