A 3kg block ‘A ‘ moving with 4 m/sec on a smooth table collides inelastically and head on with an 8kg block ‘B’ moving with speed 1.5 m/sec towards ‘A ‘. Given e = ½
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
V A = +2m/s, V B = + 
,
6Ns, 12 Ns, 33J 
Sol. using momentum conservation
e = 
3 × 4 – 8 × 1.5 = 3V 1 + 8V 2
12 – 12 = 3 V 1 + 8V 2 ∴ 3 V 1 + 8V 2 = 0 ....
coffecient of restitution 
= 
....(2)
V 2 – V 1 = 
....(2)
reat in (i) 3 × v 1 + 8 
= 0
3 × v 1 + 22 + 8 × v 1 = 0 ∴ V 1 = – 
= – 2 m/sec
∴ V 2 = – 
V 1 = – 
× (–2) = 
m/sec
applying momentum conservation eqn.
m 1 V 1 + m 2 V 2 = (m 1 + m 2 )V ∴ V = 0 so
| P D | = | m 1 ( 
– V 1 ) | = | m 1 V 1 | = 3 × 4 = 12 Ns
| J R | = |e. J D | = 6 Ns
P.E = 
mv 1 2 + 
m 2 v 2 2 – 
(m 1 + m 2 ) V 2 .
= 
× 3 × 4 2 + 
× 8 × (1.5) 2 – 0 = 33 J
Δ K = Ki – K f = 33 – 
= 
J
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems