8. FORCE AND LAWS OF MOTION
hard

$100\, g$ और $200\, g$ द्रव्यमान की दो वस्तुएँ एक ही रेखा के अनुदिश एक ही दिशा में क्रमश: $2\, m s ^{-1}$ और $1\, m\, s ^{-1}$ के वेग से गति कर रही हैं। दोनों वस्तुएँ टकरा जाती हैं। टक्कर के पश्चात् प्रथम वस्तु का वेग $1.67\, m\, s ^{-1}$ हो जाता है, तो दूसरी वस्तु का वेग ज्ञात करें। ($m/s$ में)

A

$1.125$

B

$1.556$

C

$1.365$

D

$1.165$

Solution

Mass of one of the objects, $m_1 = 100 \,g = 0.1\, kg$ 

Mass of the other object, $m_2 = 200 \,g = 0.2 \,kg$ 

Velocity of $m_1$ before collision, $v_1 = 2\, m/s$

Velocity of $m_2$ before collision, $v_2 = 1\, m/s$

Velocity of $m_1$ after collision, $v_3 = 1.67 \,m/s$

Velocity of $m_2$ after collision $= v_4$

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

$m_1v_1 + m_2v_2 = m_1v_3 + m_2v_4$

$0.1\times 2 + 0.2\times 1 = 0.1\times 1.67 + 0.2\times v_4$

$0.4 = 0.67 + 0.2\times v_4$

$v_4 = 1.165\, m/s$

Hence, the velocity of the second object becomes $1.165\, m/s$ after the collision.

Standard 9
Science

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.