AP EAPCET 05-Jul-2022 Shift 1 Paper

For ball A,
mA=1‌kg,vA=4ms−1
for ball B,mB=3‌kg,vB=0
∴ Total momentum before collision,
pi‌=mAvA+mBvB
‌=1×4+3×0=4‌kg−ms−1
Since, after collision, both body stick together, therefore, it is the case of perfectly enelastic collision. Let v be the common velocity, then total momentum after collision,
pf‌=(mA+mB)v
‌=(l+3)v=4v
According to law of conservation of linear momentum,
‌‌‌‌‌pi=pf
‌⇒‌‌4=4v⇒v=1ms−1
‌‌ Time of impact, ‌∆t=0.1 s
‌‌ Force exerted on the body B, ‌
‌‌‌‌‌FB=‌

∆pB

∆t

=‌

mB(vB−uB)

∆t

=‌

3(v−0)

0.1

‌⇒⋅FB=‌

3(1−0)

0.1

=‌

3

0.1

=30N