A wooden block of mass 1 kg travelling in a straight line with a velocity of 10 ms−1 collides with and sticks to a stationary concrete block of mass 5 kg. After the collision they move together in the same straight line. Find the velocity of the combined object.
A wooden block of mass 1 kg travelling in a straight line with a velocity of 10 ms−1 collides with and sticks to a stationary concrete block of mass 5 kg. After the collision they move together in the same straight line. Find the velocity of the combined object.
The correct option is A 1.67 ms−1
Given: mass of wooden block m1=1 kg, initial velocity of wooden block u1=10 ms−1, mass of concrete block m2=5 kg, initial velocity of concrete block u2=0.
Let v be the combined velocity of the blocks.
Initial momentum of wooden block =m1u1
Initial momentum of concrete block =m2u2
Total momentum before collision =m1u1+m2u2
⇒(1×10)+(5×0)=10 kgms−1
Total momentum after collision =(m1+m2)×v=6v kgms−1.
From the law of conservation of momentum,
Initial momentum = Final momentum
6v=10⇒v=1.67 ms−1
Therefore, velocity of the combined objectv=1.67ms−1
Given: mass of wooden block m1=1 kg, initial velocity of wooden block u1=10 ms−1, mass of concrete block m2=5 kg, initial velocity of concrete block u2=0.
Let v be the combined velocity of the blocks.
Initial momentum of wooden block =m1u1
Initial momentum of concrete block =m2u2
Total momentum before collision =m1u1+m2u2
⇒(1×10)+(5×0)=10 kgms−1
Total momentum after collision =(m1+m2)×v=6v kgms−1.
From the law of conservation of momentum,
Initial momentum = Final momentum
6v=10⇒v=1.67 ms−1
Therefore, velocity of the combined objectv=1.67ms−1
