please answer this question by solving.... please
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By momentum conservation
2mv′=√2mV
v′=v/√2
KE=2{1/2mv^2}+1/2(2m)v^2/2
=mv^2+mv^2/2
=3/2mv^2
2mv′=√2mV
v′=v/√2
KE=2{1/2mv^2}+1/2(2m)v^2/2
=mv^2+mv^2/2
=3/2mv^2
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let after explosion two identical blocks move in x & y dirⁿ respectively .. then, net momentum before collision was zero.. so net momentum after collision must be equal to zero.. as the two object started moving in perpendicular directions & momentum is vector . so resultant velocity will be √2v in the resultant direction of x & y axis . so .let the speed of 2m be "u"
2mu=√2mv.. and it gives you.. 'u'=v/√2.. and velocity's are known. now u can find increment in kinetic energy
2mu=√2mv.. and it gives you.. 'u'=v/√2.. and velocity's are known. now u can find increment in kinetic energy
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