Question $3(15$ pts$)$

The figure shows a bullet of mass $m$ moving with speed $v$ towards a stationary block $($with larger mass $M)$ that is attached to a relaxed spring, with spring constant $k.$ It collides with the block and becomes embedded in it$,$ and then the spring becomes compressed. There is no friction between the block and the surface.

$($a$)(5$ pts$)$ Which of the following formulas relates the initial speed of the bullet before the collision $(v)$ to the speed of the block and bullet immediately after the collision $(v_f)?$

$($i$)mv=m{v}_f.$

$($ii$)\frac{1}{2}m{v}^2=\frac{1}{2}(m+M)v_f^2.$

$($IIi$)nv=(m+M)v_f.$

$($iv$)\frac{1}{2}m{v}^2=\frac{1}{2}M{v}_j^2.$

CLM $...$completelly inelastic collition

$($b$)(5$ pts $)$ The total kinetic energy of the system before the collision is $K_i.$ The total kinetic energy immediately after the collision is $K_f.$ Which expression correctly relates $K_i$ and $K_f$ to the potentinl energy of the spring, $U_a,$ when it is maximally compressed after the collision?

$($i$)K_i=K_f=U_0$

$($ii$)K_i>K_f>U_0$

$($iii$)K_i>K_f=U_0$

$K_f>K_i>U_i{K}_f=U_s$cdots$P_{\text{b}\text{r}\text{e}\text{e}\text{t}}=$ elastic for boncing! $K_f(inelastic$ collision$)$

$(iv)K_f>K_i>U_i$

$(v)$ None $of$ the above.

$K_f=U_s$cdots block$-$bullet momentarilly

Sops when umpressed $(lanctic$ energy trausferted $to$ inerace $in$ opring peacros$)$

$(c)(5pts)$ Now imagine a different situation, $in$ which the bullet collides elastically with the block. $It$ will therefore bounce backwards. The maximum compression $of$ the spring would $be$

$(1)$ Smaller than $in$ the previous case.

$(ii)$ Larger than $in$ the previous case.

$(iii)$ The same $asin$ the previous case.

$P_{\text{b}\text{r}\text{e}\text{e}\text{t}}=$ elastic for boncing!

$(iv)$ Possibly larger $or$ smaller than $in$ the previous case.

Question $3(15$ pts$)$

The figure shows a bullet of mass $m$ moving with speed $v$ towards a stationary block $($with larger mass $M)$ that is attached to a relaxed spring, with spring constant $k.$ It collides with the block and becomes embedded in it$,$ and then the spring becomes compressed. There is no friction between the block and the surface.

$($a$)(5$ pts$)$ Which of the following formulas relates the initial speed of the bullet before the collision $(v)$ to the speed of the block and bullet immediately after the collision $(v_f)?$

$($i$)mv=m{v}_f.$

$($ii$)\frac{1}{2}m{v}^2=\frac{1}{2}(m+M)v_f^2.$

$($IIi$)nv=(m+M)v_f.$

$($iv$)\frac{1}{2}m{v}^2=\frac{1}{2}M{v}_j^2.$

CLM $...$completelly inelastic collition

$($b$)(5$ pts $)$ The total kinetic energy of the system before the collision is $K_i.$ The total kinetic energy immediately after the collision is $K_f.$ Which expression correctly relates $K_i$ and $K_f$ to the potentinl energy of the spring, $U_a,$ when it is maximally compressed after the collision?

$($i$)K_i=K_f=U_0$

$($ii$)K_i>K_f>U_0$

$($iii$)K_i>K_f=U_0$

$K_f>K_i>U_i{K}_f=U_s$cdots$P_{\text{b}\text{r}\text{e}\text{e}\text{t}}=$ elastic for boncing! $K_f(inelastic$ collision$)$

$(iv)K_f>K_i>U_i$

$(v)$ None $of$ the above.

$K_f=U_s$cdots block$-$bullet momentarilly

Sops when umpressed $(lanctic$ energy trausferted $to$ inerace $in$ opring peacros$)$

$(c)(5pts)$ Now imagine a different situation, $in$ which the bullet collides elastically with the block. $It$ will therefore bounce backwards. The maximum compression $of$ the spring would $be$

$(1)$ Smaller than $in$ the previous case.

$(ii)$ Larger than $in$ the previous case.

$(iii)$ The same $asin$ the previous case.

$P_{\text{b}\text{r}\text{e}\text{e}\text{t}}=$ elastic for boncing!

$(iv)$ Possibly larger $or$ smaller than $in$ the previous case.