Question

A bullet traveling at 250 m/s hits a wooden block initially at rest and is embedded in it. The speed of the block just after the bullet is embedded is 7 m/s. What is the ratio of the mass of the block to the mass of the bullet?

This question was answered before, here is the explanation: -How did this person get to the step, 243/7 * m_b = m?

By momentum conservetion:- moi = (mtm)u to S5bu zqum mbX250= (mb+m)7 Bullet mzmass of Block 2독 S0 mb= mb+m 243 243 34.71 qui

Answer 1

mb= mass of bullet

m= mass of block

v1= initial velocity of bullet

v= velocity of bullet+ block

By conservation of momentum

mb v1= (mb+m) v

mb* 250= (mb+m)*7

(250/7)mb= mb+m

250/7= (mb+m)/mb = (mb/mb)+(m/mb)= 1+( m/mb)

250/7=1 +(m/mb)

m/ mb =( 250/7)-1= (250-7)/7= 243/7= 34.71