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 prime and v 2 prime at time 2 t0 while still moving in air. The value of |(m1 v 1 prime+m2 v 2 prime)-(m1 v 1+m2 v 2)| is

Question

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 prime and v 2 prime at time 2 t0 while still moving in air. The value of |(m1 v 1 prime+m2 v 2 prime)-(m1 v 1+m2 v 2)| is (A) zero (B) (m1+m2) g t0 (C) 2(m1+m2) g t0 (D) (1/2)(m1+m2) g t0

Tardigrade Admin · Accepted Answer

|(m1 v 1 prime+m2 v 2 prime)-(m1 v 1+m2 v 2)| = mid change in momentum of the two particles mid = mid External force on the system mid × time interval =(m1+m2) g(2 t0)=2(m1+m2) g t0

zero

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

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

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

$∣ (m_{1} v_{1}^{'} + m_{2} v_{2}^{'}) - (m_{1} v_{1} + m_{2} v_{2}) ∣$
$=∣$ change in momentum of the two particles $∣$
$=∣$ External force on the system $∣ \times$ time interval
$= (m_{1} + m_{2}) g (2 t_{0}) = 2 (m_{1} + m_{2}) g t_{0}$

zero

(m1​+m2​)gt0​

2(m1​+m2​)gt0​

21​(m1​+m2​)gt0​

∣(m1​v1′​+m2​v2′​)−(m1​v1​+m2​v2​)∣ =∣ change in momentum of the two particles ∣ =∣ External force on the system ∣× time interval =(m1​+m2​)g(2t0​)=2(m1​+m2​)gt0​

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

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

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

$∣ (m_{1} v_{1}^{'} + m_{2} v_{2}^{'}) - (m_{1} v_{1} + m_{2} v_{2}) ∣$
$=∣$ change in momentum of the two particles $∣$
$=∣$ External force on the system $∣ \times$ time interval
$= (m_{1} + m_{2}) g (2 t_{0}) = 2 (m_{1} + m_{2}) g t_{0}$