Physics momentum exercise?

2015-08-30 4:23 pm

1. A bullet of 0.2 kg travelling horizontally with a speed of 300 ms-1 hits a stationary wooden block of 5kg on a smooth table and is then embedded into it . They travel with the same speed after collision which lasts for 0.05s.

(a) find the final common speed.
(b) Find the depth that the bullet penetrates into the block.

2. A particle of mass 1kg travelling with speed 4ms-1 is collided by another particle of mass 0.2 kg travelling in the same direction with speed 5ms-1. If the 0.2 kg particle reverses direction with speed of 1 ms-1 after collision and the collision time is 0.1 s.

(a) Find the speed of the 1kg particle.
(b) Calculate the force acting on the 1kg particle during collision.

回答 (1)

麻辣

2015-08-30 9:33 pm

1. m=0.2 kg travelling horizontally with v=300 m/s hits a stationary

wooden block M=5kg on a smooth table and is then embedded into it. They

travel with the same speed after collision which lasts for t=0.05s.

(a) V = ?

Momentum conservative law:

m*v = (m+M)*V

V = m*v/(m+M)

= 0.2*300/5.2

= 60/5.2

= 11.54 m/s

(b) Find the depth that the bullet penetrates into the block.

Energy conservative law: x=?

m*v^2/2 = F*x + (m+M)*V^2/2

=> F*x = [m*v^2 - (m+M)*V^2]/2

=> x = [m*v^2 - (m+M)*V^2]/2F

= [m*v^2 - (m+M)*V^2]/(2*m*a)

= [m*v^2 - (m+M)*V^2]/[2*m*(v-0)/t]

= [m*v^2 - (m+M)*V^2]*t/(2*m*v)

= [0.2*300^2 - 5.2*(60/5.2)^2]*0.05/(2*0.2*300)

= 7.21 m

2. M=1kg travelling with speed V1=4m/s is collided by m=0.2kg in the

same direction with v1=5m/s. If the m particle speed of v2=-1m/s after

collision and the t=0.1s.

(a) V2 = ?

Momentum conservative law:

m*v1 + M*V1 = m*v2 + M*V2

V2 = [m*(v1-v2) + M*V1]/M

= [0.2*(5+1) + 1*4]/1

= 0.2*6 + 4

= 5.2 m/s

(b) F = ? for M

F = M*a

= M*(V2 - V1)/t

= 1*(5.2 - 4)/0.1

= 1.2*10

= 12 N

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