Q-1.A body of 2 Kg mass having velocity 3 m/s collides with a body of 1 kg mass moving with a velocity of 4 m/s in the opposite direction. After collision both bodies stick together and move with a common velocity. Find the velocity?
Solution: System=m1v1-m2v2m1+m2=2✕3-1✕43=23m/s
Q-2.A particle with mass m and velocity v collides with a stationary particle of mass 2m, and they stick together upon impact. What is the resulting speed of the combined mass after the collision?
Solution: On applying conservation of momentum
Pi=Pf
m✕v-2m✕0=(m+2m)VSystem
mv=3mVSystem
VSystem=v3m/s
Q-3.A bullet of mass m travelling with a velocity u strikes a stationary block of mass m on a smooth surface. The collision is completely inelastic. We need to determine the common velocity after the collision and calculate the fractional loss in kinetic energy.
Solution:
On applying momentum conservation,
mu+M ✕ 0=M+mv
v=mum+M
Loss in K.E,K=12Mmm+Mu-02
Incident K.E,Ki=12mu2
Fractional Loss in K.E,KKI=MM+m
Q-4.A cart A, with a mass of 50 kg, is traveling at a speed of 20 km/h when it collides with a lighter cart B, which has a mass of 20 kg and is moving towards cart A at a speed of 10 km/h. After the collision, the two carts stick together. We need to find the speed of the combined mass after the collision.
Solution:
This is an example of an inelastic collision since the carts are moving toward each other, meaning their momenta have opposite signs. Let the common speed after the collision be V. By using principle of conservation of momentum, we can express this relationship as follows,
50 ✕ 20-20 ✕ 10=70 ✕ V
V=807km/h
Q-5.After a perfectly inelastic collision between two similar balls moving at the same speed but in different directions, the speed of the total mass is half of the initial speed. We need to determine the angle between the two balls before the collision.
Solution: Conservation of Momentum is used and is the required angle
p2=p12+p22+2p1p2 Cos
2mv22=mv2+mv2+2mvmvCos
1=1+1+2 Cos
Cos=-12
=120°
(Session 2025 - 26)