Question

Three particle of masses 20g,30g,40g are initially moving along the positive direction of the three cordinate axes respectively with the same velocity of 20 cm/ sec. When due to their mutual interaction ,the first particle comes at rest ,second acquire a velocity of (10i+20k) cm/sec. Then what is velocity of third particle?

Answer 1

we have to use , law of conservation of linear momentum.
$m_1u_1+m_2u_2+m_3u_3=m_1v_1+m_2v_2+m_3v_3$

here , m1 = 20g , m2 = 30g and m3 = 40g
u1 = 20i , u2 = 20j and u3 = 20k
v1 = 0 , v2 = (10i + 20k) and v3 = ?

20 × 20i + 30 × 20j + 40 × 20k = 20 × 0 + 30 × (10i + 20k) + 40 × v3

400i + 600j + 800k = 300i + 600k + 40v3

(400 - 300)i + 600j + (800 - 600)k = 40v3

100i + 600j + 200k = 40v3

v3 = (100i + 600j + 200k)/40

v3 =(2.5i + 15j + 5k)

hence, velocity of 3rd particle is (2.5i + 15j + 5k)

Answer 2

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we have to use , law of conservation of linear momentum.

here , m1 = 20g , m2 = 30g and m3 = 40g

u1 = 20i , u2 = 20j and u3 = 20k

v1 = 0 , v2 = (10i + 20k) and v3 = ?

20 × 20i + 30 × 20j + 40 × 20k = 20 × 0 + 30 × (10i + 20k) + 40 × v3

400i + 600j + 800k = 300i + 600k + 40v3

(400 - 300)i + 600j + (800 - 600)k = 40v3

100i + 600j + 200k = 40v3

v3 = (100i + 600j + 200k)/40

v3 =(2.5i + 15j + 5k)

hence, velocity of 3rd particle is (2.5i + 15j + 5k)