BITSAT 2018 Slot 2 Solved Paper

Given, m=5kg,v=20ms−1,θ=60°
Vertical component of velocity, vy=v‌sin‌60°
=20×

√3

2

=10√3ms−1
Time taken to reach the highest point = Time taken to reach the ground from highest point.
t=

v‌sin‌θ

g

=

vy

g

=

10√3

9.8

=1.77s
If the highest point, m splits up into two parts of masses m1=1kg and m2=4kg.
If their velocities v1 and v2 respectively, then applying the principle of conservation of linear momentum, we get
m1v1+m2v2=mv‌cos‌θ
v1+v2=5×20 ×

1

2

‌‌[∵θ=60°]
v1+4v2=5×10=50 .......(i)
Initial KE=

1

2

m(v‌cos‌θ)2
=

1

2

×5×(10)2=250J
Final KE=2 (initial KE)=2×250=500J
∴

1

2

m1v12+

1

2

m2v22=500
or

1

2

×1×v12+

1

2

×4×v22=500
orv12+4v22=1000 ........(ii)
Solving Eqs. (i) and (ii), we get
v1=30m∕s,v2=5m∕s
Hence, the separation between the two fragments
=(v1−v2)×t=(30−5)×1.77m =44.25m