The process that uses a half-life in its computation is
1 answer:
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Answer:
Part a)
v = 16.52 m/s
Part b)
v = 7.47 m/s
Explanation:
Part a)
(a) when the large-mass object is the one moving initially
So here we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
since this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
Part b)
(b) when the small-mass object is the one moving initially
here also we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
Again this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity =
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.