From conservation of linear momentum,

mv=(4m+m)vf
vf=5v
From conservation of mechanical energy,
21mv2=21(4m+m)vf2+mgh
v2=5vf2+2gh
v2=5×(5v)2+2gh
h=5g2v2
JEE Main 2019 — Physics Mechanics
A wedge of mass M=4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:
Held on 9 Apr 2019 · Verified 6 Jul 2026.
gv2
2gv2
5g2v2
7g2v2
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Two wires as shown in the figure below, made of steel and have breaking stress of $12 \times 10^8$ N/m$^2$. Area of cross-section of upper wire is $0.008$ cm$^2$ and of lower wire is $0.004$ cm$^2$. The maximum mass that can be added to pan without breaking any wire is _____ kg. (take $g = 10$ m/s$^2$) 
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