Respuesta :
does the spit ball stick to the carton once it hits? assuming this, we solve this problem using conservation of momentum
the momentum of the spitball/carton system before collision is the same as the momentum after collision
momentum before collision = 0.0031kgxV where V is the initial speed of the spitball
momentum after collision = 0.0231kgx0.28m/s= 0.006468kgm/s
(note that the mass after collision is the sum of the carton and spitball)
solving for V gives:
V= 0.006468kgm/s / 0.0031kg= 2.09 m/s
answer
his 3.1 g spit ball should hit the carton at 2.09 m/s
the momentum of the spitball/carton system before collision is the same as the momentum after collision
momentum before collision = 0.0031kgxV where V is the initial speed of the spitball
momentum after collision = 0.0231kgx0.28m/s= 0.006468kgm/s
(note that the mass after collision is the sum of the carton and spitball)
solving for V gives:
V= 0.006468kgm/s / 0.0031kg= 2.09 m/s
answer
his 3.1 g spit ball should hit the carton at 2.09 m/s
we answer this problem using conservation of momentum
the momentum of the spitball/carton system before collision is the alike as the momentum after the collision
momentum before collision = 0.0031 kg x V, the v here is the initial speed
momentum after collision = 0.0221kg x 0.28m/s = 0.0062 kg m/s
(note that the mass after collision is the sum of the carton and spitball)
solving for V gives:
V = 0.0062 kg m/s / 0.0221 kg = 0.280542986 m / s
the momentum of the spitball/carton system before collision is the alike as the momentum after the collision
momentum before collision = 0.0031 kg x V, the v here is the initial speed
momentum after collision = 0.0221kg x 0.28m/s = 0.0062 kg m/s
(note that the mass after collision is the sum of the carton and spitball)
solving for V gives:
V = 0.0062 kg m/s / 0.0221 kg = 0.280542986 m / s