A cart of mass 20 kg is moving with a velocity 5 m/sec on a straight track. a body of mass 5 kg falls vertically on the cart and both are moving together with a speed of ........ m/sec.
Answers
Let their velocities after the collision be v1
and v2. As we know for elastic collision.
Relative velocity of approach = relative velocity of separation
10−4=v2−v1⇒6=v2−v1
⇒v1=v2−6
Applying conservation of momentum,
10×10+5×4=10v1+5v2
120=10v1+5v2
120=10(v2−6)+5v2=15v2−60
15v2=180⇒v2=12cm/sec
v1=6cm/sec
Given:-
- Mass of 1st Cart ,m1 = 20kg
- Initial velocity of 1st Cart ,u1 = 5m/s
- Mass of 2nd Cart ,m2 = 5 Kg
- Initial velocity of 2nd Cart ,u2 = 0m/s
To Find:-
- Combined velocity of 1st & 2nd Cart ,v
Solution:-
Here we use the Formula of Conservation of Momentum .
We have to calculate the Final combined velocity of both cart . Therefore, Combined velocity of both cart will be
• v = m1u1+m2u2/m1+m2
where,
v is the combined velocity
m1 mass of 1st Cart
u1 initial velocity of 1st Cart
m2 is mass of 2nd Cart
u2 is initial velocity of 2nd Cart
Substitute the value we get
→ v = 20×5+5×0/20+5
→ v = 100 +0/25
→ v = 100/25
→ v = 4m/s
Therefore, the combined velocity of both cart is 4 m/s .