Two bobs P and Q of the same mass of 0.4kg are suspended by two light inextensible strings and they are initially in contact with each other. Bob P is then displaced by an angle 60 degree to the vertical and released from rest. When bob P reaches the lowest postion, it makes a head-on collision with bob Q. The strings are both 1.6m long and the size of the bobs are negligible. 18% of the kinetic energy is lost during the collision.
Q: Find the velocity of bob Q just after collision.
Physics Q about momentum(急)
2010-09-22 2:19 am
回答 (2)
2010-09-22 4:40 am
✔ 最佳答案
Assume the gravitional acceleration g=10ms^-2P.E. of bob P when it makes an angle 60 degree to the vertical
=mgh
=0.4x10x1.6-(1.6cos60)
=3.2J
By the conservation of energy,
when bob P reaches the lowest position and collides with bob Q,
P.E of bob P=82% K.E of bob Q (∵18% of K.E. is lost during the collision)
K.E of bob Q=3.2x0.82=2.624J
∴1/2 xmv^2=2.624
Then v=√(2.624x2/0.4)=3.62ms^-1
∴ the velocity of bob Q after collision is 3.62ms^-1
2010-09-22 4:49 am
Vertical height reached by P when it is displaced 60 deg.
= [1.6 - 1.6cos(60)] m = 0.8 m
Potential energy of P = 0.4g x 0.8 J, where g is the acceleration due to gravity (taken to be 10 m/s2 for simplicity of calculation)
Hence, by conservation of energy,
(1/2)(0.4)u^2 = 0.4g x 0.8
where u is the velocity of P just before collision
hence u = 4 m/s
By conservation of momentum,
0.4 x 4 = 0.4(v1) + 0.4(v2)
where v1 and v2 are the velocities of P and Q after collision respectively
i.e. 4 = v1 + v2 ------------------- (1)
But by energy conservation,
[(1/2).(0.4).(4^2)] x (1-0.18) = (1/2).(0.4).(v1)^2 + (1/2).(0.4)(v2)^2
i.e.13.12 = (v1)^2 + ((v2)^2 ------------------- (2)
solve (1) and (2) for v1 and v2 gives v2 = 0.4 m/s (rejected) or 3.6 m/s
and v1 = 3.6 m/s (rejected) or 0.4 m/s
Hence, the velocity of Q is 3.6 m/s
= [1.6 - 1.6cos(60)] m = 0.8 m
Potential energy of P = 0.4g x 0.8 J, where g is the acceleration due to gravity (taken to be 10 m/s2 for simplicity of calculation)
Hence, by conservation of energy,
(1/2)(0.4)u^2 = 0.4g x 0.8
where u is the velocity of P just before collision
hence u = 4 m/s
By conservation of momentum,
0.4 x 4 = 0.4(v1) + 0.4(v2)
where v1 and v2 are the velocities of P and Q after collision respectively
i.e. 4 = v1 + v2 ------------------- (1)
But by energy conservation,
[(1/2).(0.4).(4^2)] x (1-0.18) = (1/2).(0.4).(v1)^2 + (1/2).(0.4)(v2)^2
i.e.13.12 = (v1)^2 + ((v2)^2 ------------------- (2)
solve (1) and (2) for v1 and v2 gives v2 = 0.4 m/s (rejected) or 3.6 m/s
and v1 = 3.6 m/s (rejected) or 0.4 m/s
Hence, the velocity of Q is 3.6 m/s
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