A buggy of mass $100\, kg$ is free to move on a frictionless horizontal track. Two men, each of mass $50\, kg$, are standing on the buggy, which is initially stationary. The men jump off the buggy with velocity $=10m/s$ relative to the buggy. In one situation, the men jump one after the other. In another situation, the men jump simultaneously. What is the ratio of the recoil velocities of the buggy in two cases?
A buggy of mass $100\, kg$ is free to move on a frictionless horizontal track. Two men, each of mass $50\, kg$, are standing on the buggy, which is initially stationary. The men jump off the buggy with velocity $=10m/s$ relative to the buggy. In one situation, the men jump one after the other. In another situation, the men jump simultaneously. What is the ratio of the recoil velocities of the buggy in two cases?
- A
$5 : 4$
- B
$5 : 3$
- C
$7 : 6$
- D
$7 : 5$
Similar Questions
One projectile moving with velocity $v$ in space, gets burst into $2$ parts of masses in the ratio $1 : 3$ . The smaller part becomes stationary. What is the velocity of the other part ?
One projectile moving with velocity $v$ in space, gets burst into $2$ parts of masses in the ratio $1 : 3$ . The smaller part becomes stationary. What is the velocity of the other part ?
The balls, having linear momenta $\vec{p}_1=\hat{p} \hat{i}$ and $\vec{p}_2=-p \hat{i}$, undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. The following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $\mathrm{p}, \mathrm{a}_1, \mathrm{a}_2, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{c}_1$ and $\mathrm{c}_2$.
$(A)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}$
$(B)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}$
$(C)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}$
$(D)$ $ \vec{p}_1^{\prime}=a_1 \hat{i}+b_1 \hat{j} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=a_2 \hat{\mathrm{i}}+b_1 \hat{\mathrm{j}}$
The balls, having linear momenta $\vec{p}_1=\hat{p} \hat{i}$ and $\vec{p}_2=-p \hat{i}$, undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. The following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $\mathrm{p}, \mathrm{a}_1, \mathrm{a}_2, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{c}_1$ and $\mathrm{c}_2$.
$(A)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}$
$(B)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}$
$(C)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}$
$(D)$ $ \vec{p}_1^{\prime}=a_1 \hat{i}+b_1 \hat{j} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=a_2 \hat{\mathrm{i}}+b_1 \hat{\mathrm{j}}$
- [IIT 2008]
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